Problem 7.25 in Rudin's Principles of Mathematical Analysis goes as follows:

Suppose $\phi$ is a continuous bounded real function in the strip defined by $0\leq x\leq 1,-\infty <y<\infty$. Prove that the initial-value problem $$y'=\phi(x,y)$$ has a solution.

Below the problem, Rudin gives a hint to the reader by outlining a process by which to obtain a solution via successive approximations by the Euler method. My solution doesn't quite follow the hint exactly, which is why I'm concerned there might be something wrong with it. I give details how my solution strays from the hint at the bottom of this post.

My Solution: Fix $n$. For $i=0,\ldots,n$, put $x_{i} = i/n$. We define $f_n:[0,1]\to R^1$ to be a piecewise linear function with $f_n(0)=c$ and $$f'_n(t)=\phi (x_i,f_n(x_i))\quad\text{if $x_i<t<x_{i+1}$.}$$ We also define $$\Delta_n(t) = f_n'(t)-\phi(t,f_n(t))$$ except at the points $x_{i}$, where $\Delta_n(t)=0$. Then $$f_n(x)=c+\int_0^x\left[\phi(t,f_n(t))+\Delta_n(t)\right]\,dt.$$ Lastly we choose $M<\infty$ such that $|\phi|\leq M$. By our definition of $f$, it follows that $|f'_n(t)|\leq M$ wherever $f'_n(t)$ exists. Note that $$\phi(t,f_n(t))+\Delta_n(t)=\begin{cases}f'_n(t) &\text{if $f$ is differentiable at $t$}\\ \phi(x_{i},f_n(x_{i})) & \text{if $t=x_{i}$}\end{cases}$$ implying that \begin{align*} \left|f_n(x)\right|&=\left|c+\int_0^x\left[\phi(t,f_n(t))+\Delta_n(t)\right]\,dt\right|\\ &\leq |c|+\int_0^x\left|\phi(t,f_n(t))+\Delta_n(t)\right|\,dt \leq |c|+M. \end{align*} Thus if we set $M_1=|c|+M$, then $M_1$ serves as a uniform bound for $\{f_n\}$. Note our bound on $f'_n$ implies that $\{f_n\}$ is uniformly Lipschitz. Indeed, if $$x\leq x_{i}<x_{i+1}<\cdots<x_{j}\leq y$$ it follows that \begin{align*} |f_n(y)-f_n(x)|&\leq |f_n(y)-f_n(x_{j})|+|f_n(x_j)-f_n(x_{j-1})|+\cdots + |f_n(x_i)-f_n(x)|\\ &\leq M(y-x_j)+M(x_j-x_{j-1})+\cdots + M(x_i-x)=M(y-x). \end{align*} Ergo we can apply the Arzelà-Ascoli theorem to obtain a uniformly convergent subsequence $\{f_{n_k}\}$. Let $f$ denote the limit of this subsequence.

Our next goal will be to show that $$\phi(t,f_{n_k}(t))\to \phi(t,f(t))$$ uniformly. Fix $\epsilon>0$. Note since $[0,1]\times [-M_1,M_1]$ is compact, $\phi$ must be uniformly continuous on this set. Choose $\delta>0$ such that $\left\lVert{(x_1,y_1)-(x_2,y_2)}\right\rVert<\delta$ implies $\left|\phi(x_1,y_1)-\phi(x_2,y_2)\right|<\epsilon$ for all $(x_1,y_1),(x_2,y_2)\in R^2$. Pick $N>0$ such that $k>N$ implies $|f(t)-f_{n_k}(t)|<\delta$ for all $t\in[0,1]$. It follows that for every $k>N$ and $t\in[0,1]$, we have $\left\lVert{(t,f(t))-(t,f_{n_k}(t))}\right\rVert<\delta$. Consequently $\left|\phi(t,f_{n_k}(t))-\phi(t,f(t))\right|<\epsilon$, and $\phi(t,f_{n_k}(t))\to \phi(t,f(t))$ uniformly on $[0,1]$.

We will next demonstrate that $\Delta_{n_k}(t)\to 0$ uniformly on $[0,1]$ by noting $$\Delta_{n_k} (t)=\phi(x_{i},f_{n_k}(x_{i}))-\phi(t,f_{n_k}(t))$$ for $t\in(x_{i},x_{i+1})$. Since $\phi(t,f_{n_k}(t))\to \phi(t,f(t))$ uniformly and $[0,1]$ is compact, it follows that $\{\phi(t,f_{n_k}(t))\}$ is equicontinuous on $[0,1]$. Fixing $\epsilon>0$, pick $\delta>0$ such that $|x-y|<\delta$ implies $|\phi(x,f_{n_k}(x))-\phi(y,f_{n_k}(y))|<\epsilon$ for every $k$ and $x,y\in[0,1]$. Finally, take any $N>0$ such that $k>N$ implies $n_k>1/\delta$. Then if $k>N$ and $t\in(x_{i},x_{i+1})$, it follows that $|x_i -t |<\delta$. Consequently $$|\Delta_{n_k} (t)|=\left|\phi(x_{i},f_{n_k}(x_{i}))-\phi(t,f_{n_k}(t))\right|<\epsilon,$$ and $\Delta_{n_k}\to 0$ uniformly on $[0,1]$. We finish the proof by noting $$\phi(t,f_{n_k}(t))+\Delta_{n_k}(t)\to \phi(t,f(t))$$ uniformly on $[0,1]$. Hence, $$f(x)=\lim_{k\to \infty} f_{n_k}(x)=c+\lim_{k\to \infty}\int_0^x [\phi(t,f_{n_k}(t))+\Delta_{n_k}(t)]\,dt = c+\int_0^x \phi(t,f(t))\,dt.$$ Since $f$ is the limit of a uniformly convergent sequence of continuous functions, it follows that $f$ itself is continuous. Consequently $\phi(t,f(t))$ is continuous. This allows us to use the Fundamental Theorem of Calculus to obtain $$f'(x)=\phi(x,f(x)).$$ Combined with the fact that $f(0)=c$, this demonstrates that $f$ solves the initial value problem.

My solution specifically strays from the hint in that I don't show $\Delta_n$ is uniformly bounded and Riemann-integrable. The hint also asks the reader to show $\Delta_n \to 0$ uniformly, as opposed to just the subsequence $\Delta_{n_k}$.

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    $\begingroup$ Your piecewise linear functions look a lot like successive approximations by the Euler method... $\endgroup$ – Karl Jul 13 at 0:26
  • $\begingroup$ @Karl mmhmm I think I should have clarified that there are some details in the hint that I didn't end up using in my proof, which I put at the bottom of the question $\endgroup$ – Hrhm Jul 13 at 1:12

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