Proof verification: Baby Rudin Chapter 7 Exercise 25 (Euler Method)

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 $$0\leq x\leq 1,-\infty . Prove that the initial-value problem $$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 $$f'_n(t)=\phi (x_i,f_n(x_i))\quad\text{if x_i We also define $$\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} 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 $$\Delta_{n_k}$$.

• Your piecewise linear functions look a lot like successive approximations by the Euler method... – Karl Jul 13 at 0:26
• @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 – Hrhm Jul 13 at 1:12