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$$ \min_{|f(x)|^2} \int_{a-b}^{a+b} x^2 |f(x)|^2 dx$$ such that

$$\int_{a-b}^{a+b}|f(x)|^2 = 1$$ $$\text{$|f(x)|$ is symmetric about $a$}$$

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    We don't have any information about differentiability of $g(x)=|f(x)|^{2}$ over $[a-b, a+b]$. I understand correctly? – 2017-02-09
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    Yes. We just know $g(x)$ has a fixed area and is symmetric. – 2017-02-09
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    What does it mean to minimize over $|f(x)||^2$? that doesn't even make sense because $x$ is the integrand. What is the variable of optimization? – 2017-02-09
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    Lets say, you have to find a symmetric function with fixed area which maxmizes/minimizes the integral. – 2017-02-10

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My idea is the following. $$\int_{a-b}^{a+b}(x-a)^{2}|f(x)|^{2}dx = \int_{a-b}^{a+b}x^{2}|f(x)|^{2}dx -2a\int_{a-b}^{a+b}x|f(x)|^{2}dx+a^{2}\int_{a-b}^{a+b}|f(x)|^{2}dx = \int_{a-b}^{a+b}x^{2}|f(x)|^{2}dx - 2a\int_{a-b}^{a+b}(x-a)|f(x)|^{2}dx- 2a^{2}\int_{a-b}^{a+b}|f(x)|^{2}dx+a^{2}\int_{a-b}^{a+b}|f(x)|^{2}dx = \int_{a-b}^{a+b}x^{2}|f(x)|^{2}dx-a^{2}$$

The last is because $\int_{a-b}^{a+b} (x-a)|f(x)|^{2}dx = 0$ (as it is odd function), and because of the condition.

So you can say that minimizing $\int_{a-b}^{a+b}(x-a)^{2}|f(x)|^{2}dx$ is the same as minimizing $\int_{a-b}^{a+b}x^{2}|f(x)|^{2}dx$.

And $\int_{a-b}^{a+b}(x-a)^{2}|f(x)|^{2}dx = 2\int_{0}^{a+b}(x-a)|f(x)|^{2}dx$.

The of course you can seek the solution for odd $f(x-a)$, or for even $f(x-a)$.

However without having differentiability, I don't know how to solve it.

I know it is at maximum hint (maybe even nonsense), but I didn't manage to fit it in comments, sorry.

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    Thanks a lot. This makes the problem a lot easier. – 2017-02-09
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    I'm sorry, there was a slight mistake. I edited. – 2017-02-09
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    Yeah I caught that. No worries. – 2017-02-10