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

Let $f(x)\geq0$ be a Riemann integrable function, and

$$\int_0^1f(x)\,\mathrm dx=1, \int_0^1xf(x)\,\mathrm dx=\frac16.$$

Find the minimum value of $\int_0^1f^2(x)\,\mathrm dx$

Cauchy-Schwarz inequality? en, I'm in trouble for $f(x)\geq0$

Thank you

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.

• There are squares that shouldn't be there (minor typos) in $\langle g,1\rangle$ and $\langle g,x\rangle$ – Gabriel Romon Jun 8 '14 at 12:44
• @G.T.R, Thanks, I corrected them, "I hope". – Omran Kouba Jun 8 '14 at 13:01
• I have barely expected that the poof can be obtained so easily like this! +1 upvote – Sangchul Lee Jun 8 '14 at 23:38

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.

• +1. Still assuming continuity (which is not given). If $f\neq 0$ at some point then it is locally linear there (by a variational argument). Then $f$ is V-shaped on $[0,1]$ with a root at the tip. A tedious computation shows that $4-8x$ indeed gives the minimum square norm under the given constraints. – WimC Jun 8 '14 at 9:21
• Yes, indeed, $4-8x$ rather than $8-4x$. Sorry for the misprint. – Vladimir Jun 8 '14 at 9:24
• Well, to be completely pedantic, my argument shows that $f(x)=a+bx$ on each set $K_\epsilon=\{f(x)>\epsilon\}$ for $\epsilon>0$ and hence on $K=\bigcup_{\epsilon>0}K_\epsilon$. Continuity is neither used nor follows from this. – Vladimir Jun 8 '14 at 9:33
• Existence of the unique $L^{2}$-minimizer is easy to prove: it is a direct consequence of Hilbert space theory. Indeed, the set $$E = \{f \in L^{2}[0, 1] : \textstyle \int_{0}^{1} f = 1, \textstyle \int_{0}^{1} xf = 1/6, f \geq 0 \}$$ is closed and convex, hence contains a unique distance-minimizer. – Sangchul Lee Jun 8 '14 at 10:21
• Yeah, indeed. But still I am not sure how we, having proved that the $L_2$-minimizer is of this form, prove that the set $K$ is actually an interval. – Vladimir Jun 8 '14 at 10:25