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Let ${f : [0, 1] \rightarrow [-1, 1] }$ is a continuous function such that ${ \int_{0}^{1} x f \left(x\right) dx =0}$

Find $f(x)$ such that ${ \int_{0}^{1} \left(x ^{2 } + \frac{1}{4} \right) f \left(x\right) dx}$ has the maximum value.

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Disregard continuity. Take $f = -1 + g$ so $0 \le g \le 2$ and you want to maximize $F = \int_0^1 (x^2+1/4) (g(x)-1)\ dx$ subject to $C = \int_0^1 x g(x) \ dx = \int_0^1 x\ dx = 1/2$. Since $x^2+1/4$ is strictly convex while $x$ is linear, if you have a bit of "mass" somewhere in $(0,1)$, you can increase $F$ while keeping $C$ the same if you by splitting it up and moving half of it to the right and the other half to the left, by equal distances. So an optimal solution must have $g$ of the form $$g(x) = \cases{2 & for $x \le a$\cr 0 & for $a < x \le b$\cr 2 & for $b < x \le 1$\cr}$$ where $0 \le a \le b \le 1$, or $$f(x) = \cases{1 & for $x \le a$\cr -1 & for $a < x \le b$\cr 1 & for $b < x \le 1$\cr}$$ This makes $C = a^2 + 1 - b^2$ so $a^2 - b^2 = -1/2$, and $F = \dfrac{2}{3} (a^3-b^3) + \dfrac{1}{2}(a-b) + \dfrac{7}{12}$. Using Lagrange multipliers I get $$ a=\frac{1}{2}\,\sqrt {\sqrt {2}-1},\ b=\frac{ 1+\sqrt {2} }{2} \sqrt {\sqrt {2}-1}$$ With continuous functions, you can't attain a maximum but you can get arbitrarily close.

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i have one question ,if function $f(x)$ is continuous, and it satisfies $f:[0, 1]-->[-1 ,1]$ ,does it mean that function takes minimum value at point $0$ and maximum at point $1$?in given interval sure – dato datuashvili Jul 20 '12 at 6:46
the continuous condition was my mistake. original problem hasn't continuous condition. – Go Jiyoung Jul 20 '12 at 7:29
@dato: No, it doesn't. – Robert Israel Jul 20 '12 at 16:41

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