Let us define the following series of functions: {{begin-eqn}} {{eqn | l = \map {y_0} x       | r = b       | c =  }} {{eqn | l = \map {y_1} x       | r = b + \int_a^x \map f {t, \map {y_0} t} \rd t       | c =  }} {{eqn | l = \map {y_2} x       | r = b + \int_a^x \map f {t, \map {y_1} t} \rd t       | c =  }} {{eqn | o = \ldots       | c =  }} {{eqn | l = \map {y_n} x       | r = b + \int_a^x \map f {t, \map {y_{n-1} } t} \rd t       | c =  }} {{end-eqn}} Denote this sequence by $\sequence {y_k}_{k \mathop \in \N_0}$. What we are going to do is prove that $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_n} x$ is the required solution. === The curve lies in the rectangle === We will show that for $a - h \le x \le a + h$, the curve $y = \map {y_n}x$ lies in the rectangle $R$. That is, that $b - k < y < b + k$. Suppose $y = \map {y_{n - 1} } x$ lies in $R$. Then: {{begin-eqn}} {{eqn | l = \size {\map {y_n} x - b}       | r = \size {\int_a^x \map f {t, \map {y_{n - 1} } t} \rd t}       | c =  }} {{eqn | o = \le       | r = M \size {x - a}       | c =  }} {{eqn | o = \le       | r = M h       | c =  }} {{eqn | o = <       | r = k       | c =  }} {{end-eqn}} Clearly $y_0$ lies in $R$, and the argument holds for $y_1$. So by induction, $y = \map {y_n} x$ lies in $R$ for all $n \in \N$. === Existence === The sequence $\sequence {y_k}_{k \mathop \in N_0}$ can be expressed as a telescoping series: :$\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^n \paren {y_{k + 1} - y_k}$ The theorem contains more variables $\paren {\set {x, y_1, y_2} }$ and parameters $\paren {\set{h, k, M, A} }$ than inequality constraints. Thus, more relations between them can be chosen without affecting the constraints. Choose $\displaystyle h = \frac A 2$. For $a \le x \le h$ we have: {{begin-eqn}} {{eqn | l = \size {\map {y_{n+1} } x - \map {y_n} x}       | r = \size {\int_a^x \map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} \rd t} }} {{eqn | o = \le       | r = \int_a^x \size {\map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} } \rd t       | c = Absolute Value of Definite Integral }} {{eqn | o = \le       | r = \int_a^x A \size {\map {y_n} t - \map {y_{n - 1} } t} \rd t       | c = {{Defof|Lipschitz Condition (Real Function)}} }} {{eqn | o = \le       | r = \int_a^x A \norm {y_n - y_{n - 1} }_\infty \rd t       | c = {{Defof|Supremum Norm}} }} {{eqn | r = A \norm {y_n - y_{n-1} }_\infty \paren {x - a} }} {{eqn | o = \le       | r = A \norm {y_n - y_{n-1} }_\infty h }} {{eqn | r = \frac 1 2 \norm {y_n - y_{n - 1} }_\infty }} {{end-eqn}} By taking supremum norm of both sides, we get: :$\displaystyle \norm {y_{n + 1} - y_n}_\infty \le \frac 1 2 \norm {y_n - y_{n - 1} }_\infty$ By induction, the inequality can be extended: :$(1): \quad \displaystyle \norm {y_{n + 1} - y_n} \le \frac 1 {2^n} \norm {y_1 - y_0}$ Therefore: {{begin-eqn}} {{eqn | l = \sum_{n = 0}^\infty \norm {y_{n + 1} - y_n}       | o = \le       | r = \norm {y_1 - y_0} \sum_{n \mathop = 0}^\infty \frac 1 {2^n} }} {{eqn | o = <       | r = \infty       | c = Sum of Infinite Geometric Progression }} {{end-eqn}} Same argument applies to $-h \le x \le a$. Hence, $\sequence {y_k}_{k \mathop \in \N_0}$ converges in $\struct {\map {\CC^1} {\size{x - a} \le h}, \norm {\cdot}_\infty}$ to $y \in \map {\CC^1} {\size {x - a} \le h}$ absolutely. Therefore, the sequence is convergent: :$\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_{n + 1} } x = x_0 + \lim_{n \mathop \to \infty} \int_a^x \map f {x, \map {y_n} x} \rd x$ To find the limit, consider the following sequence: :$\displaystyle \map {g_n} x = \map f {x, \map {y_n} x}$ The sequence $\sequence {g_n}_{n \mathop \in \N_0}$ is a sequence of partial sums $\displaystyle g_0 + \sum_{k \mathop = 0}^n \paren {g_{k + 1} - g_k}$. It follows that: {{begin-eqn}} {{eqn | l = \norm {\map {g_{k + 1} } x - \map {g_k} x}        | r = \norm {\map f {x, \map {y_{k+1} } x} - \map f {x, \map {y_k} x} }  }} {{eqn | o = \le       | r = L \norm {\map {y_{k + 1} } x - \map {y_k} x}        | c = assumption in theorem }} {{eqn | o = \le        | r = L \norm {y_{k + 1} - y_k}_\infty        | c = {{Defof|Supremum Norm}} }} {{eqn | o = \le       | r = \frac 1 {2^k} \norm {y_1 - y_0}_\infty       | c = from $(1)$ }} {{end-eqn}} So $\sequence {g_n}_{n \mathop \in \N_0}$ converges to some $g$ in $\struct {\map \CC {\size {x - a} \le h}, \norm {\, \cdot \,}_\infty}$ absolutely. It follows that: {{begin-eqn}} {{eqn | l = \map g x        | r = \lim_{n \mathop \to \infty} \map {g_n} x  }} {{eqn | r = \lim_{n \mathop \to \infty} \map f {x, \map {y_n} x} }} {{eqn | r = \map f {x, \map y x} }} {{end-eqn}} On the other hand,a  Riemann integral is a continuous mapping. From Continuous Map Preserves Convergent Sequences: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \int_a^x \map f {t, \map {y_n} t} \rd t        | r = \lim_{n \mathop \to \infty} \int_a^x \paren {\map {g_0} t + \sum_{k \mathop = 0}^{n - 1} \paren {\map {g_{k + 1} } t - \map {g_k} t} } \rd t  }} {{eqn | r = \int_a^x \map g t \rd t }} {{eqn | r = \int_a^x \map f {t, \map y t} \rd t }} {{end-eqn}} We conclude that: :$\displaystyle \map y x = y_0 + \int_a^x \map f {t, \map y t} \rd t$ where: :$\map y a = y_0 + 0 = b$ and, by Fundamental Theorem of Calculus: :$\map {y'} x = 0 + \map f {x, \map y x}$ for all $x \in \R : \size {x - a} \le h$. === Uniqueness === {{AimForCont}} that the solution to IVP is not unique. Then, for the same initial conditions there exists a non-empty subset of $R$ where solutions differ. Let $y_1, y_2$ be solutions to IVP for $x \in \R : \size {x - a} \le h$. Let $x_* := \max \set {x \in \R : \size {x - a} \le h : \map {y_1} t = \map {y_2} t, \forall t \le x }$ Then: :$\displaystyle \map {y_1} x - \map {y_1} {x_*} = \int_{x_*}^x \map {y_1'} t \rd t = \int_{x_*}^x \map {f_1} {t, \map {y_1} t} \rd t$ :$\displaystyle \map {y_2} x - \map {y_2} {x_*} = \int_{x_*}^x \map {y_2'} t \rd t = \int_{x_*}^x \map {f_2} {t, \map {y_2} t} \rd t$ After taking the difference: :$\displaystyle \map {y_1} x - \map {y_2} x = \int_{x_*}^x \paren {\map {f_1} {t, \map {y_1} t} - \map {f_2} {t, \map {y_2} t}} \rd t$ Let $N \in \R$ be such that: :$N > \max \set {1, \dfrac 1 A, \dfrac 1 {A \paren {a \mathop + h \mathop - x_*} } }$ For all cases it holds that: :$x_* + \dfrac 1 {A N} < a + h$ Let: :$\displaystyle B = \max_{t \mathop \in \closedint {x_*} {x_* \mathop + \frac 1 {A N} } } \size {\map {x_2} t - \map {x_1} t} \le 2 k$ Then $\forall x \in \closedint {x_*} {x_* + \dfrac 1 {AN}}$ we have: {{begin-eqn}} {{eqn | l = \size {\map {x_2} x - \map {x_1} x}       | r = \size {\int_{x_*}^x  \paren {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t} }} {{eqn | o = \le       | r = \int_{x_*}^x  \size {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t       | c = Absolute Value of Definite Integral }} {{eqn | o = \le       | r = \int_{x_*}^x A \size {\map {x_2} t - \map {x_1} t} \rd t       | c = {{Defof|Lipschitz Condition (Real Function)}} }} {{eqn | o = \le       | r = \int_{x_*}^x A B \rd t }} {{eqn | r = A B \paren {x - x_*} }} {{eqn | o = \le       | r = A B \paren {x_* + \frac 1 {A N} - x_*} }} {{eqn | r = \frac {A B} {A N} }} {{eqn | r = \frac B N }} {{end-eqn}} Thus:  :$\forall t \in \closedint {x_*} {x_* + \dfrac 1 {A N} } : \size {\map {x_1} t - \map {x_2} t} \le \dfrac B N$ and $B \le \dfrac B N$ or $N \le 1$.  This brings us to a contradiction. Hence our assumption that the solution to IVP is not unique was false. Hence the result, by Proof by Contradiction. {{qed}}