Q:

Show norms are equiv. on $C^1[a,b]$: $\Vert f\Vert _1=\Vert f \Vert_{\infty}+\Vert f' \Vert_{\infty},\Vert f \Vert_2=|f(a)|+\Vert f' \Vert_{\infty}$

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!

A:

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 $m=\frac1{1+b-a}$.

