The first welcome problem


Problem 1: let f : \left[ 0,1 \right]\to R be a continuous differentiable function, such that \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx\le 2\int\limits_{0}^{1}{f(x)dx}} find f if f(1)=-\frac{1}{6} 😀 Thank you!!

Solution

We know that 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=\int\limits_{0}^{1}{({{({{f}^{'}}(x))}^{2}}dx}}+2\int\limits_{0}^{1}{x{{f}^{'}}(x)dx}+\frac{1}{3}

so 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=\int\limits_{0}^{1}{({{({{f}^{'}}(x))}^{2}}dx}}+2xf(x)|_{0}^{1}-2\int\limits_{0}^{1}{f(x)dx}+\frac{1}{3}

then 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx}-2\int\limits_{0}^{1}{f(x)dx}

From this we get: \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx}-2\int\limits_{0}^{1}{f(x)dx}=0

So \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=0}

The continuity of {{f}^{'}} implies {{f}^{'}}(x)=-x. That is, f(x)=-\frac{{{x}^{2}}}{2}+a

for all x\in \left[ 0,1 \right]. From f(1)=-\frac{1}{6} we obtain a=\frac{1}{3}, that is f(x)=-\frac{{{x}^{2}}}{2}+\frac{1}{3}

Thank you 😀

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About Bunchhieng

I like math, programming and web design!!
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