Anki Dokazi

1.1.1. 1.

1.1.2. 2.

1. <2>.1
   

   \( f \ge 0 \) Dokaz: zaradi definicije funkcije \( f(x) \) v 1.

2. <2>.2 QED
   

1.1.3. 3.

1. <3>.1
   

   Case: \( S(\chi_{K_n}) = \inf_D S(\chi_{K_n}, D) \).

2. <3>.2
   

   \( 0 \le S(\chi_{K_n}, D) \le S(nf, D) = n S(f, D) \) za vsako delitev \( D \) kvadra \( K \) Dokaz: 2. ter definicije zgornje Darbouxove vsote:

   \begin{align*} S(f, D) &= \sum\limits_{i = 1}^n \sum\limits_{j = 1}^m M_{ij} \left| P_{ij} \right| && M_{ij} = \sup f(x, y) \\ S(nf, D) &= \sum\limits_{i=1}^n \sum\limits_{j=1}^m n M_{ij} \left| P_{ij} \right|\\ = n S(f, D) \end{align*}
3. <3>.3
   

   \( n \cdot S(f) = 0 \), ker je \( \int\limits_K^{} f\,\mathrm{d x} = 0 \) Dokaz: predpostavka v trditvi

4. <3>.4 QED
   

1.1.4. 4.

1.1.5. 5.

1.1.6. 6.

1.1.7. 7.

1.1.8. 8.

2.1.1. <1>1

2.1.2. <1>2

1. <2>1
   

   Assume: delitvi \( D_1 = \left\{ I_i = [x_{i-1}, x_i]; \ i = 1, \ldots, m \right\} \) za \( I \) in \( D_2 = \{J_j = [y_{j - 1}, y_j]; \ j = 1, \ldots n \} \) za \( J \). Prove: \( D_1 \times D_2 \) je delitev za \( I \times J \)

   1. <3>1
      

      \( D_1 \) je projekcija na \( I \)

   2. <3>2
      

      \( D_2 \) je projekcija \( J \)

   3. <3>3 QED
      

      By: <3>1 in <3>2

2. <2>2
   

   Assume: \( P_{ij} = I_i \times J_j \) in

   \( m_{ij} (f) = \inf_{P_{ij}} f \) in \( M_{ij} (f) = \sup_{P_{ij}} f \)

   Prove: \[ s(f, D) \le \sum\limits_{i = 1}^m \inf_{x \in I_i} g(x) \left| I_i \right| = s(g, D_1) \]

   Dokaz:

   1. <3>1
      

      \[ s(f, D) = \sum\limits_{i = 1, j=1}^{m, n} m_{ij} (f) \left| P_{ij} \right| = \sum\limits_{i=1}^m \left( \sum\limits_{j = 1}^n m_{ij}(f) \left| J_j \right| \right) \left| I_i \right| \] Dokaz: definicija spodnje Darbouxove vsote ter \( \left| P_{ij} \right| = \left| J_j \right| \left| I_i \right|\)

   2. <3>2
      

      \[ \sum\limits_{j =1}^{n} m_{ij} (f) \left| J_j \right| \le \sum\limits_{j = 1}^m m_j (f(x, \cdot)) \left| J_j \right| = s(f(x, \cdot), D_2) \le \int\limits_J^{} f(x, y) \,\mathrm{d y} = g(x) \]

      Dokaz: Integrabilno po Darbouxu ter definicije \( m_{ij} \) in \( m_j \).

   3. <3>3
      

      \[ \sum\limits_{j=1}^n m_{ij} (f) \left| J_j \right| \le \inf_{x \in I_i} g(x) \]

      Dokaz: <3>2 in \( \inf_{x \in I_i} g(x) \le g(x) \)

   4. <3>2 QED
      

      Dokaz: <3>1, in <3>3 ter definicija spodnje Darbouxove vsote

3. <2>3
   

   Simetrično s supremumi

4. <2>4
   

   \( s(f, D) \le s(g, D_1) \le S(g, D_1) \le S(f, D) \) Dokaz: <2>2 in <2>3

5. <2>5
   

   Assume: \(\forall \epsilon > 0 \) Prove: \[ S(g, D_1) - s(g, D_1) < \epsilon \] Dokaz:

   1. <3>1
      

      \[ S(f, D) - s(f, D) < \epsilon \] Dokaz: \( f \) je integrabilna in obstaja taka delitev \( D \)

   2. <3>2 QED
      

      Dokaz: <3>1 in <2>4

6. <2>6 QED
   

2.1.3. <1>3 QED

3.1.1. <1> 1

3.1.2. <1>2

1. <2>1
   

   Točke nezveznosti funkcije \( \tilde{f} \) so vsebovane v \( \Gamma_{\alpha} \cup \Gamma_{\beta} \).

2. <2>2
   

   \( V(\Gamma_{\alpha} \cup \Gamma_{\beta}) = 0 \) Dokaz: Prostornina unije grafov s prostornino 0 je 0.

3. <2>3
   

   \( \tilde{f}\) je zvezna skoraj povsod Dokaz: <2>2 ter definicija s.p.

4. <2>4 QED
   

   Dokaz: <2>3 in Lebesgue

3.1.3. <1>3

3.1.4. <1>4

17.1.1. <1>1

17.1.2. <1>2

17.1.3. <1>3 QED

19.1.1. <1>1

19.1.2. <1>2 Rolleov izrek

1. Rolleov izrek
   

   Spomni se: Rolleov izrek (Matematika 1, Mrčun, str. 163)

   Predpostavke:

   • \( a, b \in \mathbb{R}, \ a < b \)
   • \( f: [a, b] \to \mathbb{R} \) zvezna funkcija
   • \( f \) odvedljiva na \( (a, b) \)

   Če velja \( f(a) = f(b) \), potem obstaja takšna točka \( c \in (a, b) \)

   \[ f'(c) = 0 \]

   V našem primeru je \( f(t) = \left\langle \gamma (t) - \gamma(t_0), w \right\rangle \)

   1. Dokaz
      
      1. <2>1
         

         \( f \) omejena.

         Dokaz: \( f \) je definirana na zaprti omejeni podmnožici in zvezna.

      2. <2>2
         

         Case: \( u, v \in [a, b], \ f(u) = \sup (f), \ f(v) = \inf(f) \)

         Prove: \( c = u \)

         Dokaz: \( u \in (a, b) \) je ekstrem funkcije \( f \), po Case <2>2

      3. <2>3
         

         \( c = v \) Dokaz: \( v \in (a, b) \) je ekstrem funkcije f , po Case <2>2

      4. <2>4
         

         \( \forall c \in (a, b) \)

         Dokaz: \( f = \text{konst} \) za \( \left\{ u, v \right\} \subset \left\{ a, b \right\} \)

      5. <2>5 QED
         

         Dokaz: <2>2, <2>3, <2>4

19.1.3. <1>3

19.1.4. <1>4

19.1.5. <1>5

19.1.6. <1>6

1. <2>1
   

   Obstaja \( \delta > 0 \), da za poljubna \( \xi_1, \xi_3 \in (t_0 - \delta, t_0 + \delta) \) Dokaz: \( \dot{\gamma}, \ddot{\gamma} \) sta zvezni v točki \( t_0 \).

2. <2>2 QED
   

   Pogoj velja tudi za \( \xi_1, \xi_3 \) za \( t_1, t_2 \in (t_0, t_0 + \delta) \)

19.1.7. <1>7

19.1.8. <1>8

19.1.9. <1>9

19.1.10. <1>10 QED

20.1.1. <1>1

20.1.2. <1>2

1. <2>1
   

   Case 1: do predznaka natančno

   \[ \vec{n} (t_1, t_2) = \frac{\left[ \gamma(t_1) - \gamma(t_2) \right] \times \left[ \gamma(t_2) - \gamma(t_1) \right]}{\left| \left[ \gamma(t_1) - \gamma(t_2) \right] \times \left[ \gamma(t_2) - \gamma(t_1) \right] \right|} \]

   Prove: \( \vec{n}(t_1, t_2) \) je zvezna funkcija argumentov \( t_1, t_2 \in \mathbb{R} ^2 \) na \( \left( t_0 - \delta, t_0 + \delta \right) \times \left( t_0 - \delta, t_0 + \delta \right) \).

   Dokaz: \( \gamma \) je zvezna funkcija argumentov \( t_2, t_1 \) iz predpostavke 1.

2. <2>2
   

   Case 2: \( \epsilon > 0 \) Case 3: definiramo \( f(t) = \left\langle \gamma(t) - \gamma(t_0), \vec{n} (t_1, t_2) \right\rangle \).

   Obstajata \( t_0 < \xi_1 < t_1 \xi_2 < t_2 \) tako, da velja

   \begin{align*} 0 &= f'(\xi_1) = \left\langle \dot{\gamma}(\xi_1), \vec{n}(t_1, t_2) \right\rangle \\ 0 &= f'(\xi_2) = \left\langle \dot{\gamma} (\xi_2), \vec{n} (t_1, t_2) \right\rangle \end{align*}

   Dokaz: Rolleov izrek

3. <2>3
   

   Obstaja \( \xi_3 \in \left( \xi_1, \xi_2 \right) \) tako, da velja

   \[ f''(\xi_3) = \left\langle \ddot{\gamma}(\xi_3), \vec{n}(t_1, t_2) \right\rangle = 0 \]

   Dokaz: Rolleov izrek

4. <2>4
   

   Velja

   \[ \left| v_0 \times \vec{n}(t_1, t_2) \right| \le \left( \left| \dot{\gamma}(t_0 - \dot{\gamma}(\xi_1)) \right| \left| \ddot{\gamma})(t_0) \right| + \left| \ddot{\gamma} (t_0) - \ddot{\gamma}(\xi_3) \right| \left| \dot{\gamma}(t_0) \right|\right) \left| \vec{n}(t_1, t_2) \right| \]

   in \( \left| \vec{n}(t_1, t_2) \right| = 1 \)

   Dokaz:

   1. <3>1
      
      \begin{align*} v_0 &= \left[ \dot{\gamma}(t_0) \times \ddot{\gamma}(t_0) \right] \times \vec{n} (t_1, t_2) \\ &= \left\langle \dot{\gamma}(t_0), \vec{n}(t_1, t_2) \right\rangle \cdot \ddot{\gamma} (t_0) - \left\langle \ddot{\gamma}(t_0), \vec{n}(t_1, t_2) \right\rangle \cdot \dot{\gamma}(t_0) \\ &= \left\langle \dot{\gamma}(t_0) - \dot{\gamma}(\xi_1), \vec{n} (t_1, t_2) \right\rangle \cdot \ddot{\gamma}(t_0) - \left\langle \ddot{\gamma}(t_0) - \ddot{\gamma}(\xi_3), \vec{n}(t_1, t_2) \right\rangle \cdot \dot{\gamma} (t_0) \end{align*}

      Dokaz:

      \( (a \times b) \times c = (a \cdot c) \cdot b - (b \cdot c) \cdot a \) ter v zadnji vrstici <2>2, ter <2>3

   2. <3>2 QED
      

      <3>1 in trikotniška neenakost

5. <2>5
   

   Case 4: \( \delta > 0 \)

   Za \( \xi_1, \xi_3 \in (t_0 - \delta, t_0 + \delta) \) velja

   \begin{align*} \left| \dot{\gamma}(t_0) - \dot{\gamma}(\xi_1) \right| &< \epsilon \\ \left| \ddot{\gamma}(t_0) - \ddot{\gamma}(\xi_3) \right| &< \epsilon \end{align*}

   Dokaz: definicija zveznosti funkcije

6. <2>6
   

   \[ \sin \phi (t_1, t_2) \le \frac{\left| \dot{\gamma}(t_0) \right| + \left| \ddot{\gamma}(t_0) \right|}{\left| \dot{\gamma} \times \ddot{\gamma} \right|(t_0)} \epsilon \]

   Kot \( \phi(t_1, t_2) \) je oster, saj je \( \epsilon > 0 \), ampak majhen.

   Dokaz:

   1. <3>1
      

      \[ \left| v_0 \times \vec{n}(t_1, t_2) \right| = \left| v_0 \right| \sin \phi \]

      Dokaz: definicija vektorskega produkta in \( \left| \vec{n}(t_1, t_2) \right| = 1 \)

   2. <3>2
      

      \[ \left| v_0 \right| \sin \phi \le \frac{\left| \dot{\gamma}(t_0) \right| + \left| \ddot{\gamma}(t_0) \right|}{\left| \dot{\gamma} \times \ddot{\gamma} \right| (t_0)} \]

      Dokaz: <3>1 in <2>4

   3. <3>3 QED
      

      Dokaz: <3>2

7. <2>7
   

   Case 5: \( \tilde{v}_0 = \frac{v_0}{\left| v_0 \right|} \)

   \[ \left| \tilde{v}_0 - \vec{n} \right| ^2 = 2(1 - \cos \phi) \]

   Dokaz: analiza 1

   \[ \left| \tilde{v}_0 - \vec{n} \right| = \left| \tilde{v}_0 \right| + \left| \vec{n} \right| - 2 \left\langle \tilde{v}_0, \vec{n} \right\rangle = 2 - 2\cos \phi \]

   in \( \left| \tilde{v}_0 \right| = \left| \vec{n} \right| = 1 \).

8. <2>8 QED
   

   Dokaz: <2>7: razlika \( 1 - \cos \phi \) gre proti 0, ko \( t_1, t_2 \to t_0 \), ker oster kot \( \phi \to 0 \).

20.1.3. <1>3 QED

22.1.1. <1>1

22.1.2. <1>2

1. <2>1
   

   \[ \gamma_i (\tilde{t}) = \gamma_i (t_0) + \dot{\gamma}_i (t_0) \cdot h + \ddot{\gamma}_i (t_0) \cdot \frac{ h ^2}{2} + o(h ^2), \quad i = 1, 2, 3 \]

   Dokaz: Razvoj \( \gamma_i (\tilde{t}) \) v Taylorjevo vrsto

2. <2>2
   
   \begin{align*} \left| \left\langle \dot{\gamma}(t_0) \cdot h ^2 + \ddot{\gamma}(t_0) \cdot \frac{h ^2}{2} + \vec{o}(h ^2), \vec{n} \right\rangle \right| &= \left| \left\langle o(h ^2), \vec{n} \right\rangle \right| \\ & \le \left| o(h ^2) \right| = o(h ^2) \end{align*}

   Dokaz: Združimo <1>1 in <2>1, trikotniška neenakost in pa skalarni produkt pravokotnih vektorjev (\( \dot{\gamma} \), \( \ddot{\gamma} \) na \( \vec{n} = \dot{\gamma} \times \ddot{\gamma} \)) je enak 0

3. <2>3 QED
   

22.1.3. <1>3 QED

27.1.1. Dokaz

1. <1>1
   

   Case 1: \( h(x) = f(x) (b - a) - x (f(b) - f(a)) \), zvezna na \( [a, b] \), odvedljiva na \( (a, b) \)

   Velja

   \begin{align*} h(a) &= f(a) (b - a) - a(f(b) - f(a)) = f(a) b - f(b) a \\ h(b) &= f(b) (b - a) - b(f(b) - f(a)) = - a f(b) + b f(a) \end{align*}

   Dokaz: Case 1

2. <1>2
   

   \[ 0 = h'(c) = f'(c) (b - a) - (f(b) - f(a)) \]

   Dokaz: <1>1 velja Rolleov izrek, \( h'(x) \) obstaja po Case 1

3. <1>3 QED
   

   Dokaz: <1>2

27.2.1. <1>1

27.2.2. <1>2

27.2.3. <1>3

27.2.4. <1>4

29.1.1. <1>1

29.1.2. <1>2

29.1.3. <1>3 QED

40.1.1. <1>1

1. <2>1
   

   \[ U(\vec{\xi}) = u \left( X (\vec{\xi}), Y(\vec{\xi}), Z(\vec{\xi}) \right) = u \left( \vec{r}(\vec{\xi}) \right) \]

   Dokaz: Case 1

2. <2>2
   

   \[ \frac{\partial U}{\partial \xi_j} = \frac{\partial u}{\partial x} \left( \vec{r}(\vec{\xi}) \right) \cdot \frac{\partial X}{\partial \xi_j} \left( \vec{\xi} \right) + \frac{\partial u}{\partial y} \left( \vec{r}(\vec{\xi}) \right) \cdot \frac{\partial Y}{\partial \xi_j} \left( \vec{\xi} \right) + \frac{\partial u}{\partial z} \left( \vec{r}(\vec{\xi}) \right) \cdot \frac{\partial Z}{\partial \xi_j} \left( \vec{\xi} \right) \]

   Dokaz: <2>1 in verižno pravilo odvajanja

3. <2>3 QED
   

   Dokaz: <2>2 se lahko zapiše kot

   \[ \frac{\partial U}{\partial \xi_j} = \left\langle \left( \nabla_{\vec{x}} u \right) \left( \vec{r}(\vec{\xi}) \right), r_j \left( \vec{\xi} \right) \right\rangle \]

40.1.2. <1>2

40.1.3. <1>3

40.1.4. <1>4

40.1.5. <1>5 QED

41.1.1. <1>1

41.1.2. <1>2 QED

42.1.1. <1>1

42.1.2. <1>2

42.1.3. <1>3 QED

43.1.1. <1>1

43.1.2. <1>2

43.1.3. <1>3 QED

44.1.1. <1>1

44.1.2. <1>2 QED

46.1.1. <1>1

46.1.2. <1>2

46.1.3. <1>3

46.1.4. <1>4 QED

48.1.1. <1>1

50.1.1. <1>1

50.1.2. <1>2

50.1.3. <1>3

50.1.4. <1>4

50.1.5. <1>5 QED

51.1.1. <1>1

1. <2>1
   

   Case 1: \( \Gamma \) sklenjena krivulja v \( \Omega \) Case 2: \( A \in \Gamma \)

   \[ \oint\limits_{\Gamma}^{} \vec{F} \mathrm{d}\vec{r} = u(A) - u(A) = 0 \]

   Dokaz: Trditev 6.9

2. <2>2 QED
   

51.1.2. <1>2

1. <2>1
   

   Case 1: \( \Gamma_1 \) in \( \Gamma_2 \) sta dve krivulji, ki povezujeta točki \( A \) in \( B \) Case 2: \( \Gamma = \Gamma_1 \cup \left( - \Gamma_2 \right) \), kjer je \( -\Gamma_2 \) krivulja \( \Gamma_2 \) z nasprotno orientacijo.

   \[ \int\limits_{\Gamma}^{} \vec{F} \mathrm{d} \vec{r} = 0 \]

   Dokaz: 2)

2. <2>2
   

   \[ \int\limits_{\Gamma_1}^{} \vec{F} \mathrm{d} \vec{r} + \int\limits_{- \Gamma_2}^{} \vec{F} \mathrm{d}\vec{r} = \int\limits_{\Gamma_1}^{} \vec{F} \mathrm{d}\vec{r} - \int\limits_{\Gamma_2}^{} \vec{F} \mathrm{d} \vec{r} = 0 \]

   Dokaz: Trditev 6.5 za spremembo predznaka

3. <2>3 QED
   

   Dokaz: <2>2

   \[ \int\limits_{\Gamma_1}^{} \vec{F} \mathrm{d}\vec{r} = \int\limits_{\Gamma_2}^{} \vec{F} \mathrm{d} \vec{r} \]

51.1.3. <1>3

1. <2>1
   

   Case 1: \( \vec{F} = \left( U, V, W \right) = \nabla u \) Case 2: \( \Gamma \) je pote od \( A \) do \( B \)

   Definiramo funkcijo

   \[ u(B) = u(A) + \int\limits_{\Gamma}^{} \vec{F} \mathrm{d} \vec{r} \]

   kjer je \( u(A) \) konstanten. Dokaz: Trditev 6.9

2. <2>2
   

   \[ \nabla (u + C) = \vec{F} \]

   za konstanto \( C \in \mathbb{R} \).

   Dokaz: Case 1 in odvod po konstanti

3. <2>3
   

   \[ u(A) = 0 \]

   Dokaz: Gradient je po <2>2 in <2>1 določen do aditivne konstante natančno

4. <2>4
   

   Case 3: \( \Gamma \) je poljubna pot od \( A \) do \( T \) znotraj \( \Omega \)

   \[ u(T) = \int\limits_{\Gamma}^{} \vec{F} \mathrm{d} \vec{r} \]

   je dobro definiran.

   Dokaz: Predpostavka

5. <2>5
   

   Velja

   \[ \nabla u = \vec{F} \]

   1. <3>1
      

      Case 1: \( \Gamma_1 \) je poljubna pot od \( A \) do \( (x, y, z) \) Case 2: \( \Gamma_2 \) je poljubna pot od \( A \) do \( (x + h, y, z) \)

      \[ u_x = \lim_{h \to 0} \frac{1}{h} \left[ u(x + h, y, z) - u(x, y, z) \right] = \lim_{h \to 0} \frac{1}{h} \left[ \int\limits_{\Gamma_2}^{} \vec{F} \mathrm{d}\vec{r} - \int\limits_{\Gamma_1}^{} \vec{F} \mathrm{d} \vec{r} \right] \]

      Dokaz: Definicija odvoda funkcije \( u \) po \( x \) ter <2>4

   2. <3>2
      

      Case 3: \( \Gamma_3 \) je daljica \( \left( x, y, z \right) \) do \( \left( x + h, y, z \right) \)

      Velja \( \Gamma_2 = \Gamma_1 \cup \Gamma_3 \)

      Dokaz: \( \Gamma_2 \) je poljubna pot

   3. <3>3
      

      Case 4: parametriziramo \( \Gamma_2 \) \( \vec{r}(t) = \left( x + ht, y, z \right), t \in [0, 1] \) Case 5: \( \odv{r} = \left( h, 0, 0 \right) \)

      \[ u(x + h, y, z) - u(x, y, z) = \int\limits_{\Gamma_3}^{} \vec{F} \mathrm{d} \vec{r} = \int\limits_0^1 \vec{F}\left( \vec{r}(t) \right) \cdot \odv{r} \mathrm{dt} = \int\limits_0^1 U \left( x + h, y, z \right) h \mathrm{dt} \]

      Dokaz: <2>4, <3>2 ter Case 4 in Case 5

   4. <3>4 QED
      

      \[ u_x = \lim_{h \to 0} \frac{1}{h} \int\limits_0^1 h \cdot U (x + h, y, z) \mathrm{dt} = U(x, y, z) \]

      Dokaz: upoštevamo <3>1, <3>3, ter, da je U zvezna in posledično tudi integral.

      Analogno še za \( y \) in \( z \).

6. <2>6 QED
   

   Dokaz <2>5

51.1.4. <1>4 QED

54.1.1. <1>1

1. <2>1
   

   Case 3: daljico \( \left[ \omega_0, \omega \right] = \left[ (x_0, y_0, z_0), (x, y, z) \right] \) parametriziramo

   \[ \vec{r}(t) = \begin{bmatrix} (1 - t)x_0 + tx \\ (1 -t) y_0 + ty \\ (1 -t) z_0 + tz \end{bmatrix} = \begin{bmatrix} r_1(t) \\ r_2(t) \\ r_3(t) \end{bmatrix} \]

   Velja

   \[ \int\limits_{\left[ \omega_0, \omega \right]}^{} \vec{F} \mathrm{d} \vec{r} = \int\limits_0^1 \left\langle \vec{F}(\vec{r}(t)), \odv{r}(t) \right\rangle_{\mathbb{R} ^3} \mathrm{dt} = \int\limits_0^1 \left\langle \vec{F}\left( \vec{r}(t) \right), \omega - \omega_0 \right\rangle \mathrm{dt} \]

   Dokaz: definicija integrala vektorskega polja ter definicija parametrizacije Case 3

2. <2>2
   

   Case 4: \( \left( \nabla u \right) \left( \omega \right) = \left( u_x, u_y, u_z \right) \left( \omega \right) \)

   \[ u_x \left( \omega \right) = \int\limits_0^1 \left( \left\langle \frac{\partial }{\partial x} \left[ \vec{F}\left( \vec{r}(t) \right) \right], \omega - \omega_0 \right\rangle + \left\langle \vec{F} \left( \vec{r}(t) \right), \frac{\partial }{\partial x} \left( \omega - \omega_0 \right) \right\rangle \right) \]

   Dokaz: \( u \) je odvedljiv, ker je integrand \( \vec{F} \) po predpostavki 2 odvedljiv, ter verižno pravilo <2>1

3. <2>3
   

   Case 5: \( \vec{F} = \left( F_1, F_2, F_3 \right) \)

   Velja

   \[ \frac{\partial }{\partial x} \left[ \vec{F}(\vec{r}(t)) \right] = \left( \frac{\partial }{\partial x} \left( F_1 (\vec{r}) \right), \frac{\partial }{\partial x} F_2 \left( \vec{r}(t) \right), \frac{\partial }{\partial x} \left( F_3 \left( \vec{r}(t) \right) \right) \right) \]

   Dokaz: Odvod vektorja je odvod posamezne komponente vektorja

4. <2>4
   

   Case 6: \( F_i = F_i \left( a_i, b_i, c_i \right), \ i = 1, 2, 3 \) in \( a_i, b_i, c_i = r_1, r_2, r_3 \)

   Za \( i = 1 \) velja

   \[ \frac{\partial }{\partial x} \left( F_1 \left( \vec{r} \right) \right) = \left( \frac{\partial F_1}{\partial a} \right) \left( \vec{r}(t) \right) \cdot \frac{\partial }{\partial x} \left( r_1 (t) \right) + \underbrace{ \left( \frac{\partial F_1}{\partial b} \right) \left( \vec{r}(t) \right) \cdot \frac{\partial }{\partial x} \left( r_2 (t) \right)}_{0} + \underbrace{ \left( \frac{\partial F_1}{\partial c} \right) \left( \vec{r}(t) \right) \cdot \frac{\partial }{\partial x} \left( r_3 (t) \right)}_{0} \]

   Dokaz: Verižno pravilo, \( a = r_1, b=r_2, c = r_3 \) ter to, da po Case 3, \( r_2 \ne r_2(x) \) in \( r_3 \ne r_3(x) \)

5. <2>5
   

   \[ \frac{\partial }{\partial x} \left[ \vec{F} \left( \vec{r}(t) \right) \right] = t \left( \frac{\partial F_1}{\partial a} \left( \vec{r} \right), \frac{\partial F_2}{\partial a} \left( \vec{r} \right), \frac{\partial F_3}{\partial a} \left( \vec{r} \right) \right) \]

   Dokaz: Odvod iz <2>4 je po definiciji <2>1 Case 3 enak \( \frac{\partial }{\partial x} \left( r_1 \right) = t \) in to za vsak \( i \) \( F_i \)

6. <2>6
   

   \[ t \left( \frac{\partial F_1}{\partial a} \left( \vec{r} \right), \frac{\partial F_2}{\partial a} \left( \vec{r} \right), \frac{\partial F_3}{\partial a} \left( \vec{r} \right) \right) = t \left( \frac{\partial F_1}{\partial a} \left( \vec{r} \right), \frac{\partial F_1}{\partial b} \left( \vec{r} \right), \frac{\partial F_1}{\partial c} \left( \vec{r} \right)\right) \]

   Dokaz: iz predpostavke \( \mathrm{rot} \vec{F} = 0 \) dobimo sledeče ekvivalence:

   \begin{align*} \begin{vmatrix} i& j & k \\ \partial_a & \partial_b & \partial_c \\ F_1 & F_2 & F_3 \end{vmatrix} = \left( \partial_b F_3 - \partial_c F_2, \partial_c F_1 - \partial_a F_3, \partial_a F_2 - \partial_b F_1 \right) &= (0, 0, 0) \\ \implies \partial_c F_1 &= \partial_a F_3 \\ \implies \partial_a F_2 &= \partial_b F_1 \end{align*}
7. <2>7
   

   Case 7: uvedemo oznako \( \tilde{\nabla} = \left( \partial_a, \partial_b, \partial_c \right) \)

   \[ \left( u_x \right) \left( \omega \right) = \int\limits_0^1 \left( \left\langle t \tilde{\nabla} F_1 \left( \vec{r} \right), \omega - \omega_0 \right\rangle + F_1 \left( \vec{r} \right) \right) \,\mathrm{d x} \]

   Dokaz: <2>1 in notacija uvedena v <2>7 Case 7, definiciji \( \omega, \omega_0 \)

8. <2>8
   

   \[ \int\limits_0^1 \left[ t \frac{\mathrm{d} \left( F_1 \circ \vec{r} \right)}{\mathrm{d} t} + \left( F_1 \circ \vec{r} \right) \right] \,\mathrm{d t} \]

   Dokaz: \( \omega - \omega_0 = \dot{r} \), in skupaj z verižnim pravilom

9. <2>9
   

   \[ \int\limits_0^1 \frac{\mathrm{d} \left( t \cdot \left( F_1 \circ \vec{r} \right) \right)}{\mathrm{d} t} \,\mathrm{d t} = \left. \left( t \cdot F_1 (\vec{r}(t)) \right) \right|_{t = 0}^{t=1} = F_1 \left( \vec{r}(1) \right) = F_1 \left( \omega \right) \]

   Dokaz: osnovni izrek analize ter definicija \( \vec{r} \) iz <2>1 Case 3

10. <2>10 QED
    

    Dokazali smo, da velja \( u_x = F_1 \) in enako velja za \( u_y = F_2, \ u_z = F_3\)

54.1.2. <1>2 QED

57.1.1. <1>1

57.1.2. <1>2

1. <2>1
   

   Case 4: \( \Omega = \Gamma_f + \Gamma_g + \mathrm{plasc} \)

   \[ \mathrm{LS} = \iint\limits_{\Gamma_f}^{} ZN_3 \,\mathrm{d S} + \iint\limits_{\Gamma_g}^{} ZN_3 \,\mathrm{d S} + \underbrace{\iint\limits_{\mathrm{plasc}}^{} ZN_3 \,\mathrm{d S }}_{0} \]

   Dokaz: Integral po množici je vsota integralov podmnožic, ki jo sestavljajo. Po predpostavki 3 \( \vec{N} \) kaže navzen in za plašč to pomeni \( \vec{N} = \left( \cdot, \cdot, 0 \right) \)

2. <2>2
   

   Case 5: parametrizacija \( \Gamma_f \) je \( \vec{r} (x, y) = \left( x, y, f(x, y) \right); \ (x, y) \in D \subset \mathbb{R} ^2 \) Case 6: parametrizacija \( \Gamma_g \) je \( \vec{R}(x, y) = \left( x, y, g(x, y) \right); (x, y) \in D \subset \mathbb{R} ^2 \)

   \[ LS = \iint\limits_D^{} Z(x, y, f(x,y)) \cdot \left( - \frac{1}{\sqrt{1 + f_x ^2 + f_y ^2}} \right) \cdot \sqrt{1 + f_x ^2 + f_y ^2} \,\mathrm{d x dy} + \iint\limits_D^{} Z(x, y, g(x, y)) \cdot \left( \frac{1}{\sqrt{1 + g_x ^2 + g_y ^2}} \right) \cdot \sqrt{1 + g_x ^2 + g_y ^2} \,\mathrm{dxdy } \]

   Dokaz:

   1. <3>1
      

      \[ \vec{r}_x \times \vec{r}_y = (- f_x, -f_y, -1) \]

      in

      \[ \vec{R}_x \times \vec{R}_y = \left( - g_x, -g_y, 1 \right) \]

      Dokaz: odvod in vektorski produkt parametrizacije Case 5 za funkciji \( f \) in \( g \) in predpostavka 3

   2. <3>2
      

      \[ \vec{N} = \frac{\vec{r}_x \times \vec{r}_y}{\left| {\vec{r}_x \times \vec{r}_y} \right|} \implies N_3 = - \frac{1}{\sqrt{1 + f_x ^2 + f_y ^2}} \]

      Dokaz:

      Vektorski produkt in <3>1

   3. <3>3 QED
      

      Dokaz: <3>2 ter uvedba parametrizacije v integralu Definicija 4.8

3. <2>3
   

   \[ \iint\limits_D^{} \left[ Z(x, y, g(x, y)) - Z(x, y, f(x, y)) \right] \,\mathrm{d xdy} = \iint\limits_D^{} \int\limits_{f(x,y)}^{g(x, y)} Z_z(x, y, t) \,\mathrm{d t} \,\mathrm{d x dy} \]

   Dokaz: <2>2 in osnovni izrek analize

4. <2>4 QED
   

   \[ \iint\limits_D^{} \int\limits_{f(x,y)}^{g(x, y)} Z_z(x, y, t) \,\mathrm{d t} \,\mathrm{d x dy} = \iiint\limits_{\Omega}^{} Z_z \,\mathrm{d V} \]

57.1.3. <1>3 QED

58.1.1. <1>1

58.1.2. <1>2

1. <2>1
   

   Case 3: \( \underline{R} = \begin{bmatrix} 0& -1 \\ 1 & 0 \end{bmatrix} \)

   \[ \vec{T} = \underline{R} \vec{N} \]

   Dokaz: \( \underline{R} \in \mathbb{R} ^2 \) rotira \( \vec{N} \) za kot \( \frac{\pi}{2} \), kar pa je ravno vektor \( \vec{T} \).

2. <2>2
   

   \[ \vec{N} = R^{-1} \vec{T} \]

   Dokaz:

   Množenje z \( \underline{R}^{-1} \)

3. <2>3
   

   \[ \left\langle \vec{G}, \vec{N} \right\rangle = \left\langle \vec{G}, \underline{R}^{-1} \vec{T} \right\rangle \]

   Dokaz:

   Upoštevamo <2>2

4. <2>4
   

   \[ \left\langle \vec{G}, \underline{R}^{-1} \vec{T} \right\rangle = \left\langle \underline{R} \vec{G}, \underline{T} \right\rangle \]

   Dokaz:

   1. <3>1
      

      Matrika \( \underline{R} \) je ortogonalna

      Dokaz: Za ortogonalno matriko velja

      \[ \underline{R}^T = \underline{R}^{-1} \iff \ \left( \underline{R}^{-1} \right)^T = \underline{R} \iff \ \underline{R}^T \underline{R} = \underline{R} \underline{R}^T = I \]

   2. <3>2
      
      \begin{align*} \left. \underline{R} \cdot \right/ \quad \left\langle \vec{G}, \underline{R}^{-1} \vec{T} \right\rangle &= \left\langle \underline{R} \vec{G}, \underline{R} \cdot \underline{R}^{-1} \vec{T} \right\rangle \\ &= \left\langle \underline{R} \vec{G}, \vec{T} \right\rangle \end{align*}

      Dokaz: Matrika je ortogonalna

   3. <3>3 QED
      
5. <2>5 QED
   

   \[ \underline{R} \vec{G} = H = (-V, U) \]

58.1.3. <1>3

58.1.4. <1>4

58.1.5. <1>5

58.1.6. <1>6 QED

59.1.1. <1>1

59.1.2. <1>2

59.1.3. <1>3

59.1.4. <1>4 QED

60.1.1. <1>1

60.1.2. <1>2

60.1.3. <1>3

60.1.4. <1>4 QED

61.1.1. <1>1 Dokaz druge Greenove identitete

61.1.2. <1>2

61.1.3. <1>3 Dokaz prve Greenove identitete

61.1.4. <1>4

61.1.5. <1>5

61.1.6. <1>6

61.1.7. <1>7

1. <2>1
   

   Vir:Directional derivati (slo. smerni odvod) Case 1: točka \( \vec{v} = (x_0, y_0, z_0) \in \mathbb{R} ^3 \) Case 2: \( u(x, y, z): \mathbb{R} ^2 \to \mathbb{R} \) zvezna realna funkcija, definirana v okolici točke \( \vec{v} \) Case 3: \( \vec{a} = (a, b, c) \in \mathbb{R} ^3 \) enotski vektor, \( \left| a \right| = 1 \)

   Definiramo smerni odvod funkcije \( u \) v točki \( \vec{v} \) v smeri vektorja \( \vec{a} \):

   \[ \frac{\partial u}{\partial \vec{a}} \left( \vec{v} \right) = \lim_{h \to 0} \frac{u \left( \vec{v} + h \vec{a} \right) - u(\vec{v})}{h} = \lim_{h \to 0} \frac{u(x + ha, y + bh, z + ch)}{h} \]

2. <2>2
   

   Smerni odvod definiramo tudi kot

   \[ \frac{\partial u}{\partial \vec{a}} \left( \vec{v} \right) = \left\langle \nabla u, \vec{a} \right\rangle \] Dokaz:

   1. <3>1
      

      Case 4: \( x_0, y_0, z_0, a, b, c\) so konstante Case 5: funkcija ene spremenljivke \( g(t) = u(x_0 + ta, y_0 + tb, z_0 + tc) = u(x, y, z) \)

      Velja

      \[ g'(t) = \lim_{h \to 0} \frac{g(t + h) - g(t)}{h} \]

   2. <3>2
      

      Za \( t = 0 \) velja

      \[ g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{u(x_0 + ah, y_0 + bh, z_0 + ch) - u(x_0, y_0, z_0)}{h} = \frac{\partial u}{\partial \vec{a}} (x_0, y_0, z_0) \]

      Dokaz: definicija odvoda ene spremenljivke <3>1, definicija Case 5, ter <2>1

   3. <3>3
      

      Velja

      \[ g'(t) = \frac{\mathrm{d} }{\mathrm{d} t} g = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial u}{\partial z} \cdot \frac{\partial z}{\partial t} = u_x \cdot a + u_y \cdot b + u_z \cdot c \]

      Dokaz: Case 5 in verižno pravilo

   4. <3>4
      

      Za \( t = 0 \) velja

      \[ g'(0) = u_x (x_0, y_0, z_0) \cdot a + u_y (x_0, y_0, z_0) \cdot b + u_z (x_0, y_0, z_0) \cdot c \]

      Dokaz: \( t = 0 \) za <3>3

   5. <3>5 QED
      

      \[ \frac{\partial u}{\partial \vec{a}} \left( \vec{v} \right) = \left\langle \nabla u, \vec{a} \right\rangle \]

      Dokaz: Izenačimo <3>4 in <3>2

3. <2>3 QED
   

   Dokaz: <2>2

61.1.8. <1>8 QED

66.1.1. <1>1

1. <2>1
   

   \[ y_1 + y_2 \in R - y_0 \iff y_0 + y_1 + y_2 \in R \]

   Dokaz:

   1. <3>1
      

      \[ L \left( y_1 + y_2 + y_0 \right) = L \left( \left( y_0 + y_1 \right) + \left( y_0 + y_2 \right) - y_0 \right) \]

   2. <3>2
      

      \[ L \left( \left( y_0 + y_1 \right) + \left( y_0 + y_2 \right) - y_0 \right) = L \left( y_0 + y_1 \right) + L \left( y_0 + y_2 \right) - L \left( y_0 \right) \]

      Dokaz: Operator \( L \) je linearen.

   3. <3>3
      

      \[ L \left( y_0 + y_1 \right) + L \left( y_0 + y_2 \right) - L \left( y_0 \right) = q + q - q \]

      Dokaz: Case 2 se lahko spremeni v \( y_1 + y_0, y_2 + y_0 \in R \)

2. <2>2 QED
   

66.1.2. <1>2

1. <2>1
   

   Case 4: \( \lambda \in \mathbb{R} \)

   \[ \lambda y_0 \in R \]

   Dokaz: ker je \( y_0 \in R \)

2. <2>2
   

   \[ \lambda y_1 \in R - y_0 \]

   Dokaz:

   \[ L(\lambda y_1) = \lambda Ly_1 \]

3. QED
   

66.1.3. <1>3 QED

74.1.1. <1>1

74.1.2. <1>2

74.1.3. <1>3

80.3.1. Trditev 7.20

1. Dokaz
   

   Dokazujemo injektivnost v obe smeri.

   1. <1>1
      

      Predpostavka 1: \( y \) reši enačbo \ref{eq:6}

      Velja:

      \[ \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} = \vec{N}(y) \]

      Dokaz: Definicija prehoda \( z_j = y^{(j - 1)} \) velja

      \begin{align*} z_1 ' &= y' = z_2 \\ z_2 ' &= y'' = z_3 \\ & \vdots \\ z_{n-1}' &= y^{(n -1)} = z_n \\ z_n ' &= y^{(n)} = f(x, y, y', \ldots, y^{(n -1)}) = f(x, z_1, \ldots, z_n) \end{align*}
   2. <1>2
      

      Predpostavka 2:

      \[ \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} = \vec{N}(y) \]

      reši sistem enačb

      \begin{align*} z_1' &= z_2 \\ z_2 ' &= z_{3} \\ z_3 ' &= z_4 \\ &\vdots \\ z_n' &= f(x, z_1, \ldots, z_n) \end{align*}

      Rešitev je injektivna.

      Dokaz:

      Velja \( y = z_1 \), \( y' = z_1' = z_2 \), itd. do \( y^{(n -1)} = z_{n-1}' = y_n \) in \( y^{(n)} = z_n ' = f(x, z_1, \ldots, z_n) = f(x, y, y', \ldots, y^{(n-1)}) \)

81.2.1. Dokaz:

1. <1>1
   

   Zapis

   \[ \int\limits_{x_0}^x y'(\xi) \,\mathrm{d }\xi = y(x) - y(x_0) = y(x) - y_0 \]

   je enak Cauchyjevemu problemu.

   Dokaz: Dobili smo ga z integriranjem CP problema, in z odvajanjem ga dobimo nazaj.

2. <1>2
   

   Obstaja natanko ena zvezna funkcija \( y: I \to \left[ y_0 - b, y_0 + b \right] \), ki reši

   \[ y(x) = y_0 + \int\limits_{x_0}^x f \left( \xi, y \left( \xi \right) \right) \,\mathrm{d } \xi \]

   Temu problemu rečemo CiP

   Dokaz:

   1. <2>1
      

      Problem CiP zadošča rešiti na zaprtem intervalu \( I\delta \).

      Opomba: \( I\delta \) je interval \( I \) skrajšan za \( \delta \) na vsaki strani.

      Dokaz: ¯\_(ツ)_/¯

   2. <2>2
      

      Case 1: Fiksiramo \( \delta (0, c) \).

      Definiramo operator \( F: y \mapsto F(y) \), kjer je \( y \) funkcija, kot

      \[ F(y):x \mapsto y_0 + \int\limits_{x_0}^x f\left( \xi, y \left( \xi \right)\right) \,\mathrm{d }\xi \]

   3. <2>3
      

      Operatorju \( F \) določimo definicijsko območje kot prostor \( \mathcal{M} = \mathcal{M}_{\delta} \) vseh zveznih funkcij \( I\delta \to \left[ y_0 - b, y_0 + b \right] \) z maksimum normo.

      1. Maksimum norma
         

         Metrika, ki poljubnima funkcijama \( \phi, \psi \in \mathcal{M} \) pripiše vrednost

         \[ d(\phi, \psi) = \max_{x \in I\delta} \left| \phi(x) - \psi(x) \right| \]

      2. Definicija metrike
         

         Naj bo \( \mathcal{N} \) poljubna množica.

         Preslikavi \( d: \mathcal{N} \times \mathcal{N} \to \mathbb{R} \) pravimo metrika na množici \( \mathcal{N} \), če za vse elemente \( x, y, z \in \mathcal{N} \) velja

         1. \( d(x, y) = d(y, x) \)
         2. \( d(x, z) \le d(x, y) + d(y, z) \)
         3. \( d(x, y) \ge 0 \)
         4. \( d(x, x) = 0 \)
         5. če \( x \ne y \), potem \( d(x, y) > 0 \)
   4. <2>4
      

      Obstaja natanko eden \( y \in \mathcal{M} \), da je

      \[ F(y) = y = y_0 + \int\limits_{x_0}^x f \left( \xi, y \left( \xi \right) \right) \,\mathrm{d } \xi \]

      tl;dr je, da iščemo fiksno točko operatorja \( F: \mathcal{M} \to \mathcal{M} \) - slika se iz \( \mathcal{M} \to \mathcal{M} \), ker je \( y \in \mathcal{M} \)

      Dokaz:

      1. Banachov izrek o fiksni točki
         

         Predpostavke:

         • \( \mathcal{N} \) poln neprazen metrični prostor
         • \( f: \mathcal{N} \to \mathcal{N} \) skrčitev

         Tedaj obstaja natanko ena točka \( b \in \mathcal{N} \), za katero velja \( f(b) = b \)

      2. <3>1
         

         Vir: https://math.stackexchange.com/questions/5015683/doubts-about-proof-that-c-a-b-with-metric-rhoxt-yt-max-limits?rq=1

         Metrični prostor \( \mathcal{M} \) je poln.

         Dokaz:

         1. Definicije
            

            Metrični prostor \( \mathcal{N} \) je poln, če je vsako Cauchyjevo zaporedje v \( \mathcal{N} \) konvergentno.

            Zaporedje \( \left( a_k \right) \) v metričnem prostoru \( \left( \mathcal{N}, d \right) \) je Cauchyjevo, če za vsako \( \epsilon > 0 \) obstaja tako velik \( N \in \mathbb{N} \), da za vsaki dve \( k,j > N; \ k, j \in \mathbb{N} \) velja \( d(a_k, a_j) < \epsilon \).

            Zaporedje v metričnem prostoru \( \mathcal{N} \) je preslikava \( \mathbb{N} \to \mathcal{N}, \ k \mapsto a_k \), ki jo običajno zapišemo kot \( \left( a_k \right)_{k \in \mathbb{N}} = \left( a_k \right) = \left( a_1, a_2, \ldots \right) \).

         2. <4>1
            

            Predpostavka 1: \( \left( \phi_n \right)_{n \in \mathbb{N}} \) je Cauchyjevo zaporedje

            Prove: Za vsak \( x \in I\delta \) je \( \left( x_n \right) \) Cauchyjevo.

            Dokaz:

            1. <5>1
               

               Za vsak \( \epsilon > 0 \) obstaja \( N \in \mathbb{N} \), da za vsaka \( j, k > N; \ j, k \in \mathbb{N} \) velja

               \[ d(\phi_n, \phi_m) < \frac{\epsilon}{2} \]

               Dokaz: Predpostavka 1, da je \( \left( \phi_n \right)_{n \in \mathbb{N}} \) Caucyjevo zaporedje.

               V katerem primeru to ne bi veljalo?

            2. <5>2
               

               Za vsak \( x \in I\delta \) velja

               \[ \left| \phi_n (x) - \phi_m (x) \right| < d(\phi_n, \phi_m) \]

               Dokaz: <5>1

            3. <5>3 QED
               

               Za vsak \( x \in I\delta \) velja

               \[ \left| \phi_n - \phi_m \right| < \frac{\epsilon}{2} \]

               Dokaz: iz <5>1 in <5>2

         3. <4>2
            

            Prove:

            Ko gre \( n \to \infty \), gre enakomerno \( \phi_n \to \phi \) na \( I_{\delta} \)

            Dokaz:

            1. <5>1
               

               Predpostavka: \( \lim_{m \to \infty} \phi_m = \phi \)

               Za vsak \( \epsilon > 0 \) obstaja \( N \in \mathbb{N} \) tako, da za vsak \( n > N \) velja

               \[ \left| \phi_n - \phi \right| = \left| \phi_n - \lim_{m \to \infty} \phi_m \right| \le \lim_{m \to \infty} \left| \phi_n - \phi_m \right| < \frac{\epsilon}{2} < \epsilon \]

               za vsak \( x \in I\delta \)

               Dokaz: <4>1 velja Opomba: Dokazali smo enakomerno konvergenco

            2. <5>2
               

               Za vsak \( I\delta \) intervala je so \( \phi \) zvezne.

               Dokaz: <5>1 drži in \( \phi_n \) so zvezne

            3. <5>3 QED
               

               Dokaz: \( \phi \in \mathcal{M} \)

         4. <4>3
            

            Funkcija \( \phi_n \) je zvezna.

            Dokaz: Predpostavka <2>3, \( \mathcal{M} \) je prostor vseh zveznih funkcij.

      3. <3>2
         

         \( F: \mathcal{M} \to \mathcal{M} \) je skrčitev.

         1. Definicija
            
            • \( \left( \mathcal{M}, d \right) \)
            • \( \left( \mathcal{M}', d' \right) \)

            Preslikava \( f: \mathcal{M} \to \mathcal{M}' \) je skrčitev, če obstaja takšna realna konstanta \( q \), da je \( 0 \le q < 1 \) in da velja

            \[ d'(f(x) , f(y)) \le q d(x, y) \]

            za vse točke \( x, y \in \mathcal{M} \)

         2. <4>1
            

            Case 2: \( \phi, \psi \in \mathcal{M} \) dve funkciji

            Velja

            \[ d \left( F \left( \phi \right), F \left( \psi \right) \right) = d \left( y_0 + \int\limits_{x_0}^x f(\xi, \phi(\xi)) \,\mathrm{d }\xi, y_0 + \int\limits_{x_0}^x f \left( \xi, \psi(\xi) \right) \,\mathrm{d }\xi \right) \]

            Dokaz: Definicija operatorja \( F \) <2>2

         3. <4>2
            

            Velja

            \[ d \left( y_0 + \int\limits_{x_0}^x f(\xi, \phi(\xi)) \,\mathrm{d }\xi, y_0 + \int\limits_{x_0}^x f \left( \xi, \psi(\xi) \right) \,\mathrm{d }\xi \right) = \max_{x \in I\delta} \left| \int\limits_{x_0}^x f \left( \xi, \phi(\xi) \right) \,\mathrm{d }\xi - \int\limits_{x_0}^{x} f \left( \xi, \psi \left( \xi \right) \right) \,\mathrm{d }\xi \right| \]

            Dokaz: Definicija metrike \( d \) <2>3

         4. <4>4
            

            \[ \max_{x \in I\delta } \left| \int\limits_{x_0}^x \left[ f \left( \xi, \phi(\xi) \right) - f \left(\xi, \psi \left( \xi \right) \right) \right] \,\mathrm{d } \xi \right| \ge \left| \int\limits_{x_0}^x \left[ f \left( \xi, \phi(\xi) \right) - f \left( \xi, \psi \left( \xi \right) \right) \right] \,\mathrm{d } \xi \right| \]

         5. <4>5
            

            \[ \left| \int\limits_{x_0}^x f \left( \xi, \phi(\xi) \right) \,\mathrm{d }\xi - \int\limits_{x_0}^{x} f \left( \xi, \psi \left( \xi \right) \right) \,\mathrm{d }\xi \right| \le \int\limits_{x_0}^x \left| f \left( \xi, \phi \left( \xi \right) \right) - f \left( \xi, \psi \left( \xi \right) \right) \right| \,\mathrm{d }\xi \]

            Dokaz: Glej integrale s parametrom

         6. <4>6
            

            Case 3: \( \gamma \in [0, 1] \)

            \[ \int\limits_{x_0}^x \left| f \left( \xi, \phi \left( \xi \right) \right) - f \left( \xi, \psi \left( \xi \right) \right) \right| \,\mathrm{d }\xi \le \int\limits_{x_0}^x \gamma \left| \phi(\xi) - \psi \left( \xi \right) \right| \,\mathrm{d }\xi \]

            Dokaz: Definicija enakomerno Lipschitzove funkcije

         7. <4>7
            

            \[ \int\limits_{x_0}^x \gamma \left| \phi - \xi \right| \,\mathrm{d } \xi \le \int\limits_{x_0}^x \gamma \max_{x \in I\delta} \left| \phi \left( \xi \right) - \psi \left( \xi \right) \right| \,\mathrm{d } \xi \]

            Dokaz: absolutna razlika bo vedno manjša ali enaka od maksimalne razlike

         8. <4>7
            

            \[ \int\limits_{x_0}^x \gamma d \left( \phi, \psi \right) \,\mathrm{d }\xi = \left( x - x_0 \right) \gamma d \left( \phi, \psi \right) \]

            Dokaz: Definicija metrike ter osnovni izrek analize

         9. <4>8
            

            \[ \gamma d \left( \phi, \psi \right) \left( x - x_0 \right) \le \gamma \left| a - \delta \right| d \left( \phi, \psi \right) \]

            Dokaz: Predpostavka 2, definicija pravokotnika \( P \), kjer ima \( x \) maksimalno vrednost \( a + x_0 \) in potem je na skrčitvi \( I\delta \) \( x_0 + a - \delta \).

         10. <4>9 QED
             

             Dokaz: Za \( \delta < a \), vendar da bo \( \delta \) dovolj blizu \( a \), bo \( \gamma \left| a - \delta \right| < 1 \) in po definicije skrčitve, bo \( F \) skrčitev.

      4. <3>3 QED
         

         Dokaz: <3>1 in <3>2

   5. <2>5 QED
      

      Pogoji za Banachov izrek so dokazani, in ga lahko uporabimo. Obstaja fiksna točka. <2>4

3. <1>3 QED
   

   Dokaz: <1>2 velja

82.1.1. <1>1

82.1.2. <1>2

82.1.3. <1>3

82.1.4. <1>4

82.1.5. <1>5

82.1.6. <1>6 QED

84.2.1. Dokaz:

1. <1>1
   

   \[ \vec{y}_p ' = \left( \underline{Y}\int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} \right) ' \]

   Dokaz: Trditev

2. <1>2
   

   \[ \left( \underline{Y} \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} \right)' = \left( \underline{Y} \right)' \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} + \underline{Y} \left( \int\limits_{}^{} \underline{Y}^{-1}\vec{b} \,\mathrm{d x}\right) ' \]

   Dokaz: odvod produkta

3. <1>3
   

   \[ \left( \underline{Y} \right)' \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} + \underline{Y} \left( \int\limits_{}^{} \underline{Y}^{-1}\vec{b} \,\mathrm{d x}\right) ' = A\underline{Y} \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} + \underline{Y} \underline{Y}^{-1} \vec{b} \]

   Dokaz:

   Velja \( \underline{Y}' = A \underline{Y} \) ter \( \left( \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} \right)' = \underline{Y}^{-1} \vec{b} \)

4. <1>4 QED
   

   \[ A\underline{Y} \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} + \underline{Y} \underline{Y}^{-1} \vec{b} = A \left( \underline{Y} \int\limits_{}^{} \underline{Y}^{-1} \vec{b} \,\mathrm{d x} \right) + \vec{b} = A \vec{y}_p + \vec{b} \]

   Dokaz: \( \underline{Y} \underline{Y}^{-1} = 1 \) ter trditev

86.1.1. <1>1

86.1.2. <1>2

86.1.3. <1>3

86.1.4. <1>4

86.1.5. <1>5 QED

89.1.1. Dokaz:

1. <1>1
   

   Case 1: \( s_n \sum\limits_{j = 0}^{n} \frac{B^j}{j!} \)

   Zaporedje \( \left( s_n \right)_n \) je Cauchyjevo.

   Spomnimo se: Za vsak \( \epsilon > 0 \) obstaja tako velik \( N \in \mathbb{N} \), da za vsaki števili \( m, n > N; \ m, n \in \mathbb{N} \) velja \( d(s_m, s_n) < \epsilon \) Dokaz:

   1. <2>1
      

      Case 2: \( m > n \)

      \[ \left\lVert s_m - s_n \right\rVert = \left\lVert \sum\limits_{j = 0}^m \frac{B^j}{j!} - \sum\limits_{j = 0}^n \frac{B^j}{j!} \right\rVert \]

      Dokaz: definicija zaporedja.

   2. <2>2
      

      \[ \left\lVert \sum\limits_{j = n + 1}^m \frac{B^j}{j!} \right\rVert \le \sum\limits_{j = n + 1}^m \frac{\left\lVert B^j \right\rVert}{j!} \]

      Dokaz: Vsoti v <2>1 se odštejeta in je tisto na levi strani, kar ostane ter neenakost

   3. <2>3
      

      \[ \sum\limits_{j = n + 1}^m \frac{\left\lVert B^j \right\rVert}{j!} \le \sum\limits_{j = n + 1}^{m} \frac{\left\lVert B \right\rVert^j}{j!} \]

      Dokaz: ponovno neenakost

   4. <2>4
      

      \[ \sum\limits_{j = n + 1}^m \frac{\left\lVert B \right\rVert^j}{j!} \]

      konvergira k 0, ko gresta \( m, n \to \infty \)

      Dokaz:

      Taylorjeva vrsta \( f(x) = e^x = \sum\limits_{j = 0}^{\infty} \frac{x^j}{j!} \) konvergira povsod na \( \mathbb{R} \). Dana vsota je številska vrsta za \( x = \left\lVert B \right\rVert \).

   5. <2>5 QED
      

      Zaradi <2>4, gre tudi <2>1 proti 0, ko gresta \( m, n \to \infty \), kar pomeni, da je vrsta Cauchyjeva.

2. <2>2 QED
   

   Dokaz: <1>1

91.1.1. <1>1

91.1.2. <1>2

91.1.3. <1>3

91.1.4. <1>4

91.1.5. <1>5

91.1.6. <1>6

91.1.7. <1>7

91.1.8. <1>8

91.1.9. <1>9

91.1.10. <1>10

91.1.11. <1>11

91.1.12. <1>12 QED

93.1.1. <1>1

93.1.2. <1>2

93.1.3. <1>3

93.1.4. <1>4

93.1.5. <1>5 QED

96.1.1. <1>1

96.1.2. <1>2

96.1.3. <1>3 QED

97.1.1. <1>1

97.1.2. <1>2

97.1.3. <1>3

97.1.4. <1>4

97.1.5. <1>5

97.1.6. <1>6

97.1.7. <1>7 QED

103.1.1. <1>1

1. <2>1
   

   \[ 0 \le \left\langle x + \lambda y, x + \lambda y \right\rangle \]

   Dokaz: Definicija 9.1, 4)

2. <2>2
   

   \[ \left\langle x + \lambda y, x + \lambda y \right\rangle = \left\langle x, x + \lambda y \right\rangle + \left\langle \lambda y, x + \lambda y \right\rangle \]

   Dokaz: Definicija 9.1, 1)

3. <2>3
   

   \[ \left\langle x, x + \lambda y \right\rangle + \left\langle \lambda y, x + \lambda y \right\rangle = \left\langle x, x + \lambda y \right\rangle + \lambda \left\langle y, x + \lambda y \right\rangle \]

   Dokaz: Definicija 9.1 2)

4. <2>4
   

   \[ \left\langle x, x + \lambda y \right\rangle + \lambda \left\langle y, x + \lambda y \right\rangle = \left\langle x + \lambda y, x \right\rangle + \lambda \left\langle x + \lambda y, x \right\rangle \]

   Dokaz: Definicija 9.1 3) in pa smo na prostoru \( \mathbb{R} \)

5. <2>5 QED
   

   \[ \left\langle x + \lambda y, x \right\rangle + \lambda \left\langle x + \lambda y, x \right\rangle = \lVert x \rVert ^2 + \lVert y \rVert ^2 + \lambda \left\langle x, y \right\rangle \]

   za vsak \( x, y \in V \) ter za vsak \( \lambda \in \mathbb{R} \) Dokaz: Definicija 9.1 in definicija norme *

103.1.2. <1>2

103.1.3. <1>3 QED

106.1.1. <1>1

106.1.2. <1>2

1. <2>1
   

   Case 1: \( a \in A \) Case 2: \( y_n \in A^{\perp } \) Case 3: \( y_n \) konvergira k \( y \)

   \[ \lim_{n \to \infty} \left| \left\langle y - y_n, a \right\rangle \right| \le \lim_{n \to \infty} \lVert y - y_n \rVert \lVert a \rVert \]

   Dokaz: CBS neenakost

2. <2>2
   

   \[ \lim_{n \to \infty} \lVert y - y_n \rVert = 0 \]

   Dokaz: Case 3

3. <2>3 QED
   

   <2>1 gre proti 0, ker gre <2>2 proti 0, torej \( y \in A^{\perp } \)

106.1.3. <1>3 QED

108.1.1. 1)

1. <1>1
   

   Case 1: \( y_1, y_2 \in Y \) Case 2: \( x - y_1 \perp Y \) Case 3: \( x - y_2 \perp Y \)

   \[ x - y_1 - x + y_2 = -y_1 + y_2 \in Y \]

   Dokaz: Odštejemo Case 2 in Case 3, ter Case 1

2. <1>2
   

   \[ y_2 - y_1 \perp Y \]

   Dokaz: Case 3 in Case 2

3. <1>3 QED
   

   Dokaz: iz <1>1 in <1>2 sledi, da je \( y_2 - y_1 \perp y_2 - y_1 \), kar je možno samo, če \( y_1 = y_2 \)

108.1.2. 2)

1. <1>1
   

   Case 1: \( z \in Y \)

   \[ x - z = \underbrace{x - P_Y x}_{\in Y^{\perp}} + \underbrace{P_Yx - z}_{\in Y} \]

   Dokaz: Prištejemo in odštejemo isto vrednost ter definicija 9.11

2. <1>2 QED
   

   \[ \lVert x - P_Y x \rVert ^2 \le \lVert x - z \rVert ^2 \]

   Dokaz: Pitagorov izrek

108.1.3. 3)

1. <1>1
   

   Case 1: \( x_1, x_2 \in X \) Case 2: \( x_1 - P_Y x_1, x_2 - P_Y x_2 \in Y^{\perp} \)

   \[ \left( x_1 + x_2 \right) - \underbrace{\left( P_Y x_1 + P_Y x_2 \right)}_{\in Y} \in Y ^{\perp} \]

2. <1>2 QED
   

   Iz 1) sledi \( P_Y x_1 + P_Y x_2 = P(x_1 + x_2) \)

108.1.4. 4)

1. <1>1
   

   Case 1: \( x = (x - P_Y x) + P_Y x \)

   \[ \lVert x \rVert ^2 = \underbrace{\lVert x - P_Y x \rVert ^2}_{\ge 0} + \lVert P_Y x \rVert \ge \lVert P_Y x \rVert \]

2. <1>2 QED
   

   Dokaz: <1>1

108.1.5. 5)

1. <1>1
   

   Case 1: \( \left( y_n \right)_n \subset Y \) Case 2: \( y_n \to x \) za neki \( x \in X \)

   \[ x \in Y \]

   1. <2>1
      

      \[ P_Y y_n \to P_Y x \]

      Dokaz: \( P \) je zvezen

   2. <2>2
      

      \[ \lVert P_Y y_n - P_Y x \rVert = \lVert P_Y \left( y_n - x \right) \rVert \]

      Dokaz: \( P \) je linearen 3)

   3. <2>3
      

      \[ \lVert P_Y \left( y_n - x \right) \rVert \le \lVert y_n - x \rVert \]

      Dokaz: 4)

   4. <2>4
      

      <2>3 je enako 0

      Dokaz: Case 2

   5. <2>5
      

      \[ P_Y y_n = y_n \]

      Dokaz: Case 1, \( y_n \in Y \)

   6. <2>6
      

      \[ P_Y x = x \]

      Dokaz:

      \[ P_Y x \Leftarrow P_Y y_n = y_n \Rightarrow x \]

   7. <2>7 QED
      

      \( x \in Y \) iz <2>6

2. <1>2 QED
   

   Prostor je linearen in zaprt

109.1.1. <1>1

109.1.2. <1>2

109.1.3. <1>3

109.1.4. <1>4 QED

111.1.1. <1>1

111.1.2. <1>2

111.1.3. <1>3