Find $f\in L^2(0,1)$ with $\int_0^1 xf(x)dx = \langle x, f(x)\rangle = 1$ of minimal norm. I would like to get more hints to the following question.
Find $f\in L^2(0,1)$ with $\int_0^1 xf(x)dx = \langle x, f(x)\rangle = 1$ of minimal norm (with the standard norm in $L^2(0,1)$).
I figured that I need to do something with orthogonality, but I have no Idea how to proceed.
I think I should decompose the space $M = \{g\in L^2(0,1)| \int_0^1 xg(x)dx= 1\}$ in a direct sum, such that $M = G\oplus G^{\perp}$. Furthermore, I could imagine this space $M$ as a hyperplane with $G^{\perp}$ sticking out and which holds the solution. So maybe I need to do something with the an inproduct being zero and maybe something with translations.
Question: is that even remotely correct and if so, how could I tie that together?
Thank you for your time.
 A: This is in the first place a problem of linear algebra: You are given a vector $a\ne0$ in a certain inner product space, and ask for the shortest vector $f$ satisfying $$\langle f,a\rangle=1\ .$$ Now the general vector $f$ satisfying this condition has the form
$$f={a\over\|a\|^2}+ v,\qquad v\perp a,$$
and the shortest of these $f$ is obviously the vector $$f_0:={a\over\|a\|^2}\ .$$
In the problem at hand the given vector is the function $a: \ x\mapsto x$ $(0\leq x\leq1)$ having $\|a\|^2={1\over3}$. It follows that the solution to your problem is the function
$$f_0(x):=3x\ .$$
A: Writing your $M$ as a direct sum doesn't make much sense, since $M$ is not a vector space.
Let $f_0=2$, so $\int_0^1 x f_0(x)=1$. Let $S$ be the space of all $g\in L^2([0,1])$ such that $\int x g(x)=0$. Then $$M=\{f_0-g:g\in S\}.$$So the function you want is $f=f_0-g_0$, where $g_0$ is the element of $S$ closest to $f_0$. The minimality implies that $$\int_0^1(f_0-g_0)g=0$$for every $g\in S$, which is to say that $$f_0-g_0\in S^\perp.$$
So what is $S^\perp$? Say $T$ is the span of the function $x$. Then $T^\perp =S$, hence $S^\perp = T$. So there exists a scalar $c$ with $$f_0(x)-g_0(x)=cx,$$
or $$g_0=f_0-cx.$$
So you only need to find $c$ so that $f_0-cx\in S$...
A: Hint: use shifted Legendre polynomials. They give an orthogonal base of $L^2(0,1)$ with respect to the usual inner product. If
$$ f(x) = c_0 + \sum_{n\geq 1} c_n\cdot P_n(2x-1) $$
we have:
$$ c_0 = \int_{0}^{1}f(x),\qquad c_1 = 3\int_{0}^{1}\left(x-\frac{1}{2}\right)f(x)\,dx $$
and by Parseval's identity:
$$ \|f\|_2^2 = \int_{0}^{1} f(x)^2\,dx = c_0^2 + \sum_{n\geq 1}\frac{c_n^2}{2n+1}, $$
so the problem boils down to finding the minimum of $c_0^2 + \frac{c_1^2}{3}$ under the constraint $\frac{c_1}{3}+\frac{c_0}{2}=1$.
