## 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 😀