$\int_0^1f(x)dx=1, \int_0^1xf(x)dx=\frac16$ minimum value of $\int_0^1f^2(x) dx$?

Solution 1:

We equip the space of Riemann square integrable functions on $[0,1]$ with the usual scalar product. Now we have $$ \int_0^1(1-2x)f(x)dx=1-\frac{1}{3}=\frac{2}{3} $$ Hence $$ \frac{2}{3}=\int_0^{1/2}(1-2x)f(x)dx+\underbrace{\int_{1/2}^1(1-2x)f(x)dx}_{\leq 0} $$ it follows that, if $g(x)=4(1-2x){\bf 1}_{[0,1/2]}$ then $$ \frac{8}{3}\leq\int_0^{1}4(1-2x){\bf 1}_{[0,1/2]}f(x)dx=\langle f,g\rangle\tag{1} $$ On the other hand we have $$\eqalign{ \Vert g\Vert^2&=16\int_0^{1/2}(1-2x)^2dx=\frac{8}{3}\leq\langle f,g\rangle\cr} $$ So $\langle f-g,g\rangle\geq0$, hence $$ \Vert f\Vert^2=\Vert f-g\Vert^2+\Vert g\Vert^2+2\langle f-g,g\rangle\geq \Vert f-g\Vert^2+\Vert g\Vert^2\tag{2} $$ But $g$ is positive and satisfies $$\langle g,1\rangle =4\int_0^{1/2}(1-2x) dx=1,\quad \langle g,x\rangle =4\int_0^{1/2}x(1-2x) dx=\frac{1}{6} $$ So, from $(2)$ we conclude that $\Vert f\Vert^2\geq\Vert g\Vert^2$ with equality, if and only if $f=g$, and the desired minimum is $\Vert g\Vert^2=8/3$.

Acknowledgment I want to acknowledge here that this solution, came from Vladimir's observation, presented in his answer.

Solution 2:

Let $f$ provide the desired minimum. If $g$ is continuous, supported in $\{x:f(x)>0\}$, and satisfies $\int g dx=\int xg dx=0$, then $f+tg$ satisfies the constraints for small $t$ and we have $\int (f+tg)^2dx\ge\int f^2 dx$ for any $t$ $\implies$ $\int fg dx=0$. This holds for every continuous $g$ satisfying $\int g dx=\int xg dx=0$ and supported in $\{x:f(x)>0\}$, and hence there exist constants $a$ and $b$ such that $f(x)=a+bx$ on the set $K=\{x:f(x)>0\}$. If we could somehow prove that, for optimal $f$, $K$ has to be an interval of the form $[0,x_0)$, then the answer probably is $f(x)=4-8x$ for $x\in[0,1/2]$ and $0$ afterwards.