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I ran across the following problem with a friend while we were studying for quals. Neither of us are really quite sure where to start. It feels like a differential equation. This is probably easy, but we were not able to get a handle on how to proceed. I wish I could tell you what I tried, but after thinking on this problem for some time, I simply do not have ideas of any real substance (other than what I mention after the problem statement).

Here is the question as it appears on the old qual:

"Let $K$ be a continuous function on the unit square $0\leq x,y\leq1$ satisfying $|K(x,y)|<1$ for all $x$ and $y$. Show that there is a continuous function $f(x)$ on $[0,1]$ such that we have

$$f(x) + \int_0^1 f(y)K(x,y)dy=\sin(x^2)$$

where $0\leq x \leq 1$. Can there be more than one such function $f$?"

I will say that I was able to show that given $K$ as it is, $\exists\,C\in(0,1)$ such that $|K|\leq C$ on the square, and that a function defined as

$$G(x)=\int_0^1 g(y)K(x,y) dy$$

will be continuous, assuming that $g$ is continuous on $[0,1]$.

Any suggestions or possible solutions would be greatly appreciated.

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HINT:

Let $f^{(n)}(x)$ be given by the recursive relationship

$$f^{(n)}(x)=\sin x^2-\int_0^1K(x,y)f^{(n-1)}(y)dy$$

with $f^{(0)}=0$. Then, show that

$$\begin{align} \left|f^{(n)}(x)-f(x)\right|&=\left|\int_0^1K(x,y)\left(f^{(n-1)}(y)-f(y)\right)dy\right|\\\\ &\le\int_0^1|K(x,y)|\left|f^{(n-1)}(y)-f(y)\right|dy\\\\ &<\lambda \left|\left|f^{(n-1)}(y)-f(y)\right|\right|_{\infty} \end{align}$$

for some $\lambda<1$ and iterate.

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  • $\begingroup$ Nice. I should have seen that. Thanks for clearing that up for me. $\endgroup$ – fxy Jul 5 '15 at 23:27
  • $\begingroup$ @fxy Thank you! And you're certainly welcome. It was my pleasure. The series solution is called The Neumann Series and is rich in both theory and application (e.g., The Born Approximation). $\endgroup$ – Mark Viola Jul 5 '15 at 23:28
  • $\begingroup$ Yea, I actually have seen Neumann series (in the context of numerical linear algebra). For some reason I get a mental block when it comes to analysis. It's like I somehow forget everything I know. Anyway, much appreciated. $\endgroup$ – fxy Jul 5 '15 at 23:33
  • $\begingroup$ Just keep practicing and you'll do well! $\endgroup$ – Mark Viola Jul 5 '15 at 23:34
  • $\begingroup$ Thanks for the encouragement. Quals scare the crap out of me. $\endgroup$ – fxy Jul 5 '15 at 23:34
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Dr. MV has covered the existence statement quite well. To get uniqueness, suppose $f$ and $g$ are both functions satisfying the conditions. Then, we would have $$f(x) - g(x) = - \int_0^1 [f(y) - g(y)]K(x,y)\;dy$$

Let $x$ be such that $|f(x) - g(x)|$ is maximal on $[0,1]$. If $f(x) \not= g(x)$, then we can divide by $f(x) - g(x)$ to obtain $$1 = - \int_0^1 \frac{f(y) - g(y)}{f(x) - g(x)} K(x,y)\; dy$$

But the term in the integrand always has absolute value less than $1$.

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  • $\begingroup$ Thanks for your answer. I would mark it right also if I could. $\endgroup$ – fxy Jul 5 '15 at 23:28
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This is just meant to be a long comment.

I just want to make you aware that this can be proved easily using the Banach fixed point theorem, which states that if $X$ is a complete metric space, then any map $\;T:X \to X$ which is Lipchitz with constant less than $1$ has a unique fixed point.

(If you're taking quals then you must be familiar with this?)

Let's see how this implies the above theorem. Let our metric space be $X:=C[0,1]$ with the sup norm, and define $T:X \to X$ as $$(Tf)(x) = \sin(x^2)-\int_0^1f(y)K(x,y)dy$$

Then $T$ is Lipchitz with constant less than $1$. Indeed, letting $\lambda := \sup_{x,y \in [0,1]} K(x,y)$, we can easily show (using the same kind of argument as in Dr. MV's answer) that $$\|Tf-Tg\|_{\infty} \leq \lambda \|f-g\|_{\infty}$$

Thus, by the Banach fixed point theorem, $T$ has a unique fixed point $f_0$, which gives both existence and uniqueness of your question at once.

Remark: The answer given by Dr. MV is essentially the essence of the proof of the Banach fixed point theorem, but I just wanted to make you aware of the statement in the more general setting.

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  • $\begingroup$ Thanks for your comment. Actually, when I was writing it up for myself, I ended up doing it much like this. His approach pointed me to this as well. Thanks again. $\endgroup$ – fxy Jul 6 '15 at 0:17
  • $\begingroup$ Yeah sure no problem :) $\endgroup$ – Shalop Jul 6 '15 at 0:20

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