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Let $f\in C^1[a,b]$ be a function. The Sobolev norm is defined as follows:

$$\|f\|_{1,2}=\left[\int_a^b \left(f(x)^2 +f'(x)^2\right)dx\right]^{1/2}$$

I want to find a constant $C$ such that $\|f\|_\infty\le C\|f\|_{1,2}$ for all $C^1[a,b]$.

How do I do so? I've tried some approaches with inequalities, but didn't succeed. Would appreciate a hint.

Also, does the existence of such a constant imply that the norms are equivalent? (Of course, we also need to find $c$ so that $\|f\|_\infty\ge c\|f\|_{1,2}$, but just curious. Since $(C^1[a,b], \|\cdot\|_\infty$ is complete and $(C^1[a,b], \|\cdot\|_{1,2})$ is incomplete, can the norms be possibly equivalent?

sequence
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    You may find some hint in this. See the answer in particular. –  Oct 26 '17 at 06:22
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    If the space is complete under one norm but not complete under another norm, both norms cannot be equivalent. – daw Oct 26 '17 at 09:49

1 Answers1

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Let $f\in E={\mathcal C}^1([a,b])$. Then there exists $m\in I=[a,b]$ such that $|f(t)|\geq |f(m)|$ for all $t\in I$. We have also $$f(x)=\int_m ^x f^{\prime}(t)dt+f(m)$$ hence $$|f(x)|\leq |f(m)|+\int_{a}^{b}|f^{\prime}(t)|dt\leq \frac{1}{b-a}\int_a^b|f(t)|dt+\int_{a}^{b}|f^{\prime}(t)|dt$$

By Cauchy-Schwarz, we have $(\int_a^b|f(t)|dt)^2\leq (b-a)\int_a^b|f(t)|^2dt$ and $(\int_{a}^{b}|f^{\prime}(t)|dt)^2\leq (b-a)\int_a^b|f^{\prime}(t)|^2dt$, and so there exists a constant $c_1>0$ such that $$|f(x)|\leq c_1((\int_a^b|f(t)|^2dt)^{1/2}+(\int_a^b|f^{\prime}(t)|^2dt)^{1/2}) $$ for all $x\in I$. To finish, use that if $u,v\geq 0$, we have $u+v\leq \sqrt{2}\sqrt{u^2+v^2}$.

Kelenner
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  • Can you please clarify how you got $|f(m)| \leq \frac{1}{b-a}\int_a^b|f(t)|dt$? Also, how did you get $\int_a^b|f(t)|dt)^2\leq (b-a)\int_a^b|f(t)|^2dt$? I think by Cauchy-Schwarz, $\int_a^b|f(t)|dt)^2\geq \int_a^b|f(t)|^2dt$, but I'm not sure how your inequality follows from this. – sequence Oct 28 '17 at 20:58
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    You have $|f(m)|\leq |f(t)|$ for all $t$. Now integrate from $a$ to $b$. The Cauchy-Scharz inequality says that $(\int_a^b |g(t)||f(t)|dt)^2\leq (\int_a^b g(t)^2dt)(\int_a^b f(t)^2dt)$, and now take $g=1$. – Kelenner Oct 29 '17 at 08:08