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}} What we are going to do is prove that $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_n} t$ is the required solution. There are five main steps, as follows: === 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 $n \in \N$. === Bounded Nature of Adjacent Differences === We will show that: :$\displaystyle \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$ This is also to be proved by induction. Suppose that this holds for $n-1$ in place of $n$. Let this be the induction hypothesis. We have: :$\displaystyle \map {y_n} x - \map {y_{n - 1} } x = \int_a^x \paren {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \rd t$ We also have that: :$\size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le A \size {\map {y_{n - 1} } t - \map {y_{n - 2} } t}$ by the Lipschitz condition. By the induction hypothesis, it follows that: $\displaystyle \size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le \frac {M A^{n - 1} \size {t - a}^{n - 1} } {\paren {n - 1}!}$ So: {{begin-eqn}} {{eqn | l = \size {\map {y_n} x - \map {y_{n - 1} } x}       | o = \le       | r = \frac {M A^{n - 1} } {\paren {n - 1}!} \size {\int_a^x \size {t - a}^{n - 1} \rd t}       | c =  }} {{eqn | r = \frac {M A^{n - 1} } {n!} \size {x - a}^n       | c =  }} {{end-eqn}} For the base case, we use $n = 1$: :$\displaystyle \size {\map {y_1} x - b} \le \size {\int_a^x \map f {t, b} \rd t} \le M \size {x - a}$ Thus by induction: :$\displaystyle \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$ for all $n$. === Uniform Convergence of Sequence === Next we show that the sequence $\sequence {\map {y_n} x}$ converges uniformly to a limit for $a - h \le x \le a + h$. From Bounded Nature of Adjacent Differences above, we have: {{begin-eqn}} {{eqn | o =        | r = b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots       | c =  }} {{eqn | o = \le       | r = b + M h + \cdots + \frac {M A^{n - 1} h^n} {n!} + \cdots       | c =  }} {{end-eqn}} From Radius of Convergence of Power Series over Factorial, it follows that $b + M h + \cdots + \dfrac {M A^{n - 1} h^n} {n!} + \cdots$ is absolutely convergent for all $h$. Hence, by the Weierstrass M-Test: :$b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots$ converges uniformly for $a - h \le x \le a + h$. Since its terms are continuous functions of $x$, its sum $\displaystyle \lim_{n \mathop \to \infty} \map {y_n} x = \map y x$ is also continuous from Combination Theorem for Sequences. === Solution Satisfies Differential Equation === We now show that $y = \map y x$ satisfies the differential equation $y' = \map f {x, y}$. Since: :$\map {y_n} x$ converges uniformly to $\map y x$ in the open interval $\openint {a - h} {a + h}$ from Uniform Convergence of Sequence above :$\size {\map f {x, y} - \map f {x, y_n} } \le A \size {y - y_n}$ from the Lipschitz condition in $y$ it follows that $\map f {x, \map {y_n} x}$ tends uniformly to $\map f {x, \map y x}$. Letting $n \to \infty$ in: :$\displaystyle \map {y_n} x = b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t$ we get: :$\displaystyle \map y x = b + \int_a^x \map f {t, \map y t} \rd t$ The integrand $\map f {t, \map y t}$ is a continuous function of $t$. Therefore the integral has the derivative $\map f {x, y}$. Also, we have that $\map y a = b$. === Uniqueness of Solution === We now show that the solution $y = \map y x$ that we have found is the ''only'' solution where $\map y a = b$. {{AimForCont}} there is another such solution, $y = \map Y x$, say. Let $\size {\map Y x - \map y x} \le B$ when $\size {x - a} \le h$. (Certainly we could take $B = 2 k$.) Then: :$\displaystyle \map Y x - \map y x = \int_a^x \paren {\map f {t, \map Y t} - \map f {t, \map y t} } \rd t$ But: :$\size {\map f {t, \map Y t} - \map f {t, \map y t} } \le A \size {\map Y t - \map y t} \le A B$ So: :$\size {\map Y t - \map y t} \le A B \size {x - a}$ Repeating the argument, we can get successive estimates for the upper bound of $\size {\map Y x - \map y x}$ in $\openint {a - h} {a + h}$. This gives: :$\displaystyle \frac {A^2 B} {2!} \size {x - a}^2, \ldots, \frac {A^n B} {n!} \size {x - a}^n, \ldots$ But this sequence tends to $0$. So $\map Y x = \map y x$ in $\openint {a - h} {a + h}$. This contradicts the supposition that $\map Y x$ and $\map y x$ are different. Hence by Proof by Contradiction it follows that $\map y x$ is unique. {{qed}}