Existence/uniqueness of a Continuous Function

Question

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.

Answer 1

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.

Answer 2

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 $1$.

Answer 3

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.