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For $\Omega=(0,1). $Prove that there exists $M>0$ such that $$||u||_{C^0(\overline{\Omega})}\le M||u||_{H^1(\Omega)}$$ for all $u\in H^1(\Omega).$

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Hi, welcome to Math.SE. We generally don't prefer questions that just say "Prove X" with no further discussion or motivation. If this is a homework problem, please follow the guidelines listed here. You can use the "edit" button to make changes. Also, there are better tags for this question than just "inequality"; I've added some. –  Nate Eldredge May 25 '13 at 12:33
    
Hi, it's not a homework. I just wonder why it's said to be easy in someone's essay. Thanks for your advice. –  anhkhoavo1210 May 25 '13 at 12:36
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1 Answer

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Assume that $u(0)=0$. Then $$ u(x)=\int_0^x u'(s)ds=\int_0^1 \textbf{1}_{(0,x)}u'(s)ds. $$ Now you use Cauchy-Schwarz inequality. $$ u(x)\leq \|u'\textbf{1}_{(0,x)}\|_{L^2}=\int_0^1 (u'(s))^2\textbf{1}_{(0,x)}ds\leq \int_0^1 (u'(s))^2 ds. $$ Right?

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Why you can assume $u(0)$ like that? I also do like your way, but don't think that. –  anhkhoavo1210 May 25 '13 at 12:41
    
In general you can't. This is a simpler case. For an unbounded domain you can do the same trick paying with an $L^2$ norm of $u$. –  guacho May 25 '13 at 12:43
    
Let me think how to deal with $u(0)$... –  guacho May 25 '13 at 12:48
    
In particularly, $M$ in someone's essay is $\sqrt{2}$. I'm strongly wondering. The problem is of $u(0)$... –  anhkhoavo1210 May 25 '13 at 13:11
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Use the fact that there is a continuous extension operator $P \colon H^1(0,1) \to H^1(\mathbb{R})$, and exploit the fact that $C_0^\infty(\mathbb{R})$ is dense in $H^1(\mathbb{R})$. See also Brezis, Analyse fonctionnelle, Théorème VIII.7. –  Siminore May 25 '13 at 13:38
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