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What is the trick, to prove $\| u\|_{L^2(\Gamma)} \leq k \frac{1}{r}\| u\|_{L^2(\Omega)} + r \| \nabla u\|_{L^2(\Omega)} $ ? $\Gamma$ is one side of $\Omega:= [0,r] \times [0,r] $. I tried partial differentiation, but it doesnt work.

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You'll probably want to add more tags to this to get it more attention. "inequality" is a bit vague and doesn't reflect the content well. –  Simon Hayward Dec 3 '12 at 15:35
    
For which $u$ do you want such an inequality? –  Davide Giraudo Dec 3 '12 at 15:45
    
For all $u \in H^1(\Omega)$. –  JamesB Dec 3 '12 at 15:59

1 Answer 1

Let $r=1$ for simplicity; the general case follows by rescaling. Also, $\Gamma=[0,1]\times \{0\}$. We may assume $u$ is continuous, by the usual density argument. Let $v(x,y)=u(x,0)$; the goal is to bound the $L^2$ norm of $v$ on the square $Q=[0,1]\times [0,1]$. To this end, if suffices to estimate $\|u-v\|_{L^2(Q)}$ by $\|\nabla u\|_{L^2(Q)}$. Well, for each $x\in [0,1]$ we have $$ \sup_{y\in[0,1]}|u(x,y)-v(x,y)| \le \int_0^1 |\nabla u(x,y)|\,dy \tag1$$ Square both sides and use Cauchy-Schwarz: $$ \sup_{y\in[0,1]}|u(x,y)-v(x,y)|^2 \le \int_0^1 |\nabla u(x,y)|^2\,dy \tag2$$ The supremum on the left majorizes the corresponding integral: $$ \int_0^1 |u(x,y)-v(x,y)|^2 \,dy \le \int_0^1 |\nabla u(x,y)|^2\,dy \tag3$$ Finally integrate in $x$ to get $\|u-v\|_{L^2(Q)}\le \|\nabla u\|_{L^2(Q)}$.

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