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The problem:

Let $a>0$ and let $g\in C^0([-a,a])$. Prove that there exists a unique function $u\in C^0([-a,a])$ such that $$u(x)=\frac x2u\left(\frac x2\right)+g(x),$$ for all $x\in[-a,a]$.

My attempt At first sight I thought to approach this problem as a fixed point problem from $C^0([-a,a])$ to $C^0([-2a,2a])$, which are both Banach spaces if equipped with the maximum norm. However i needed to define a contraction, because as it stands it is not clear wether my operator $$(Tu)(x)=\frac x2u\left(\frac x2\right)+g(x)$$ is a contraction or not. Therefore I tried to slightly modify the operator and I picked a $c>a>0$ and defined $$T_cu=\frac 1cTu.$$ $T_cu$ is in fact a contraction, hence by the contraction lemma i have for granted the existence and the uniqueness of a function $u_c\in C^0([-a,a])$, which is a fixed point for $T_cu.$ Clearly this is not what I wanted and it seems difficult to me to finish using this approach. Am I right, is all what I have done useless? And if this were the case, how to solve this problem?

Thanks in advance.

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Admittedly I did not check the details, but quite obviously your T is a contraction if $|a| < 2$. I'd start with that and try to scale if $a$ is large. – user20266 May 10 '12 at 18:43
Yes It is indeed a Contraction if $a<2$. What do you mean by scaling? – uforoboa May 10 '12 at 18:45
Notice $T:C[-a,a]\to C[-a,a]$, so if you can prove that $T^n$ (composed n-times) is a contraction, by the uniqueness of the fixed point you get a fixed point for $T$. – Jose27 May 10 '12 at 18:49
solve the problem for $a < 2$ and for $a \ge 2 $ solve the equation for $f(x) = g(\alpha x)$ with $\alpha$ such that $f: [-2+\varepsilon,2 - \varepsilon]$. Then transform back. Don't know whether it works, it's your homework :-) – user20266 May 10 '12 at 18:50
up vote 1 down vote accepted

Your approach will work if $a<2$, in the general case, write $$u(x)=\frac x2u\left(\frac x2\right)+g(x)$$ as $$u(x)=x^k\alpha_ku\left(\frac x{2^k}\right)+F_k(g)(x)$$ where $F_k$ is a functional of $g$ and $\alpha_k$ satisfy the recurrence relation $\alpha_{k+1}=\frac{\alpha_k}{2^{k+1}}$. Therefore, $a^{k+1}\alpha_k$ converges to $0$ and you apply Banach fixed point for $T_k(u)(x)=x^k\alpha_ku\left(\frac x{2^k}\right)+F_k(g)(x)$ for a $k$ large enough.

share|cite|improve this answer to solve this problem?

A direct road is to define $u$ as $$ u(x)=\sum_{n=0}^{+\infty}2^{-n(n+1)/2}x^ng(2^{-n}x), $$ and to prove that $u$ is indeed continuous on $[-a,a]$.

Hint: For every sequence of continuous functions $(g_n)_{n\geqslant0}$ such that $\|g_n\|_\infty\leqslant cA^n$ for every $n$, for some finite $c$ and $A$, the series $\sum\limits_{n=0}^{+\infty}2^{-n(n+1)/2}g_n$ defines a continuous function.

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As noted already, your approach works for $a<2$.

Now suppose that $a\geq 2$. We will first construct $u$ on [-1,1] and then double the width of this interval iteratively. Given a function $g$, the equation $$\tag{$\ast$} u(x) = \frac{x}2 u\left(\frac{x}2\right) + g(x)$$ uniquely determines a function $u_0$ on [-1,1]. Now define $u_1$ by $$u_1(x)=\begin{cases} u_0(x) &\text{if $x\in[-1,1]$} \\ \frac{x}2 u_0\left(\frac{x}2\right) + g(x) &\text{otherwise}.\end{cases}$$ We now have a function $u_1$ on [-2,2] satisfying $(\ast)$. As long as $2^{k+1}<a$, continue this way defining $$u_{k+1}(x) = \begin{cases} u_{k}(x) &\text{if $x\in[-2^{k},2^{k}]$} \\ \frac{x}2 u_{k}\left(\frac{x}2\right) + g(x) &\text{otherwise}.\end{cases}$$ At a certain point, $2^{k+1}\geq a$. Define $u_{k+1}$ only on $[-a,a]$, then stop. Note that all the $u_k$ are continuous.

It is clear that the last function you constructed is the unique solution to the problem (solution by construction, uniqueness because $u_0$ is unique and because $(\ast)$ forces all the other function values).

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