# Existence and Uniqueness of a solution

I'm stuck with this problem so hopefully somebody will help me :) Here you are:

Let $K\in C([0,2])$ be positive, decreasing and such that $K(0)=1$. Prove that for every $h\in C([0,1])$ there exists a unique solution $u\in C([0,1])$ to the equation

$$u(x)=h(x)+\int_0^1K(x+y)u(y)\mathrm d y,\qquad \forall x\in [0,1].$$

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I don't have an answer, but this kind of exercise should make you think of compactness of the operator defined by $T_h(f)(x) = h(x) + \int_0^1 K(x+y)f(y)\,\mathrm{d}y$ and some kind of fixed point theorem. – kahen Sep 2 '11 at 12:34
@Rahul... stuck means that i even don't know how to approach this problem – uforoboa Sep 2 '11 at 12:42
If $K$ is equal to $1$ on the whole of its range, then the solution is not necessarily unique (for example, if $h$ is identically zero, then $u$ can be any constant function). So K has to be strictly decreasing, right? – TonyK Sep 2 '11 at 12:51
@user15453 I'll make kahen's hint slightly more explicit and then you should really give that a good shot: Looking at your problem and kahen's hint, you want to show there exists a solution, it's unique, and also we note that it must be a fixed point of the operator kahen gave. These keywords are reminiscent of a famous theorem. Think about applying the Banach fixed point theorem to the operator kahen gave. – Ragib Zaman Sep 2 '11 at 12:54
one of my first thought was for K to be strictly decreasing, because i thought about TonyK example. On the text it is written just Decreasing... not strictly decreasing.. i don't know what is the correct interpretation to be honest – uforoboa Sep 2 '11 at 13:00

Consider the map (as defined by kahen in a comment)

$$T_h f(x) = h(x) + \int_0^1 K(x+y) f(y) dy$$

which is a linear map from $C([0,1])$ to itself. Then

$$T_h f(x) - T_h g(x) = \int_0^1 K(x+y) (f(y) - g(y)) dy$$

so

$$| T_h f(x) - T_h g(x) | \leq \int_0^1 K(x+y) |f(y) - g(y)| dy$$

where we used that $K$ is positive. Now, take $|f-g|$ in $\sup$ over $[0,1]$, and use that $K$ is strictly decreasing, so $\int_0^1 K(x+y) dy \leq \int_0^1 K(y) dy < 1$, you have that for any $h$, $T_h$ is a contraction mapping.

Now just apply the Banach fixed-point theorem and you are done.

There seems to be some confusion about whether $K$ is taken to be strictly decreasing. Note that in the crucial aspect of the proof above is that $\int_0^1 K(x+y) \leq 1-\epsilon$ for some positive $\epsilon$. That $K$ is strictly decreasing is sufficient, but not necessary.

On the other hand, TonyK indicated in his comment that in the case that $K\equiv 1$ and $h = 0$ there are infinitely many solutions (take $u$ to be any constant function). Analogous to the case of linear algebra where a change of parameter can change a linear system from having infinitely many solutions to no solutions, we see the same phenomenon here. Let $K\equiv 1$ and assume $h$ is a function with integral $\int_0^1 h(x) dx \neq 0$. Then the given integral equation admits no solutions.

Since $K \equiv 1$, the equation reduces to

$$u(x) = h(x) + \int_0^1 u(y) dy$$

Integrate both sides in $x$, you get

$$\int_0^1 u(x) dx = \int_0^1 h(x) dx + \int_0^1 dx \int_0^1 u(y) dy$$

so we can subtract the left hand side from the right hand side and get

$$0 = \int_0^1 h(x) dx \neq 0$$

which is absurd. In fact, you have the following claim: if $K\equiv 0$, the equation either has no solutions (when $h$ has non-zero average) or infinitely many solutions (when $h$ has zero average) differing from each other by constant shifts.

So by the principle of "homework exercises should be solvable", you can probably assume that whatever the source of the problem, the phrase "decreasing" is intended to mean "strictly decreasing".

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I was going to suggest iterating the mapping you've called $T_h$ and seeing if one can prove that the sequence converges (uniformly?) to a solution. But you're way ahead of me. – Michael Hardy Sep 2 '11 at 12:58
K is actually not given to be strictly decreasing, so perhaps we need to do an iteration like @Michael suggested. – Ragib Zaman Sep 2 '11 at 13:02
@Ragib: If $K$ is not strictly decreasing (that is, if $\int_0^1K(y)dy$ is allowed to equal 1), not even iteration can save you. See TonyK's comment to the main question. Also, in a lot of literature decreasing is taken to mean strictly decreasing, while the "closed" condition is called non-increasing. – Willie Wong Sep 2 '11 at 13:12
@Willie ahh it appears that's the correct interpretation of the problem then. Sorry for the mix up. – Ragib Zaman Sep 2 '11 at 13:15
(In analogue with the difference between positive and nonnegative; of course, in that case, the split is down a linguistic line: French users considers positif to mean nonnegative and positif strictement to mean positive.) – Willie Wong Sep 2 '11 at 13:16