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Here is what I got as a proof. My question is at the end. Thanks

On $ C^1[a,b]$ we have the norms $$\Vert f\Vert _1 = \Vert f \Vert_{\infty} + \Vert f' \Vert_{\infty},\quad \Vert f \Vert_2 = |f(a)| + \Vert f' \Vert_{\infty}.$$ We will show these are equivalent. Here $\Vert f \Vert_{\infty} = \sup\limits_{x\in [a,b]}|f(x)|$. In particular, by the definition of the supremum, we have $|f(a)| \leq \sup\limits_{x\in [a,b]}|f|$ so that it follows that for $M=1$ we have $$ \Vert f \Vert_2 =|f(a)| + \Vert f' \Vert_{\infty} \leq M\left( \Vert f \Vert_{\infty} + \Vert f' \Vert_{\infty} \right) = M\Vert f\Vert _1.$$ Now we want to find a number $m$ such that $m \Vert f\Vert _1 \leq \Vert f\Vert _2$. let $\beta = \inf\limits_{x\in [a,b]}|f(x)|$ and $\alpha =\sup\limits_{x\in [a,b]}|f(x)|$ and set $m= \frac{\beta }{\alpha + 1}\leq 1$. Then we have that $m \Vert f \Vert _{\infty} = m\alpha = \frac{\alpha\beta}{\alpha + 1} < \beta \leq |f(a)|$. Also, as $m\leq 1$, we see that $m\Vert f' \Vert_{\infty}\leq \Vert f' \Vert_{\infty}$ so that $m\Vert f\Vert_{1} = m\Vert f \Vert_{\infty}+m\Vert f' \Vert_{\infty} \leq |f(a)| + \Vert f' \Vert_{\infty} = \Vert f \Vert_2$. So we have found an $M$ and $m$ such that $$m\Vert f \Vert_1 \leq \Vert f \Vert _2 < M \Vert f\Vert _1.$$

My concern with this proof is that for many functions we have $\inf\limits_{x\in[a,b]}|f(x)|=0$ and thus $m=0$. I think this is a problem. Is there any way around this or can anyone see another $m$ that will work? Thanks in advance!

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    $\begingroup$ Do you mean your space is $C^1[a,b]$ ? A continuous function may have no derivative. $\endgroup$ – Stop hurting Monica Apr 30 '14 at 20:56
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    $\begingroup$ You need an $m$ that is independent of $f$, so its definition cannot really use $f$. Can you bound $\lvert f(x)\rvert$ using only $\lVert f\rVert_2$? $\endgroup$ – Daniel Fischer Apr 30 '14 at 20:56
  • $\begingroup$ You may use mean value theorem to bound $||f||_\infty$ using $|f(a)|$ and $||f'||_\infty$ ? $\endgroup$ – Stop hurting Monica Apr 30 '14 at 21:00
  • $\begingroup$ @Jean-ClaudeArbaut Yes, we may use it. Also, you are right, I did mean $C^1[a,b]$ thanks for the spot! $\endgroup$ – Slugger Apr 30 '14 at 21:08
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For any $x\in[a,b]$, the Mean Value Theorem gives you $f(x)=f(a)+f'(c)\,(x-a)$, for some $c\in[a,b]$. Then $$ \|f\|_\infty\leq|f(a)|+\|f'\|_\infty\,(b-a) $$ Letting $k=1+b-a$, $$ \|f\|_\infty+\|f'\|_\infty\leq k\,(|f(a)|+\|f'\|). $$ So you can take $m=\frac1{1+b-a}$.

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  • $\begingroup$ You are welcome! $\endgroup$ – Martin Argerami May 4 '14 at 16:45

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