# Proving that $\Vert\cdot \Vert$ defined on $C^{1}[a,b]$ is a norm

In proving that $$\Vert\cdot \Vert$$ defined on $$C^{1}[a,b]$$ by $$\Vert{f \Vert}=\max\limits_{a\leq t\leq b}\left|f(t)\right|+\max\limits_{a\leq t\leq b}\left|\dfrac{d}{dt}f(t)\right|$$ is a norm, I encountered a problem. It's getting the equation to satisfy the first condition.

MY WORK

Let $$f\in C^{1}[a,b],$$ then

\begin{align} \Vert{f \Vert}=0 &\leftrightarrow \max\limits_{a\leq t\leq b}\left|f(t)\right|+\max\limits_{a\leq t\leq b}\left|\dfrac{d}{dt}f(t)\right|=0 \\& \leftrightarrow \left|f(t)\right|+\left|\dfrac{d}{dt}f(t)\right|=0,\;\;t\in [a,b] \\& \leftrightarrow f(t)+\dfrac{d}{dt}f(t)=0,\;\;t\in [a,b]\\& \leftrightarrow f(t)=e^{-t},\;\;t\in [a,b]\end{align} I'm not getting $$f(t)=0,\;\;\forall\,t\in [a,b].$$ Please, where did I get it wrong? Can someone help me? As to the other two conditions, I have no problems with them.

• The last 2nd row. $|x|+|y|=0 \iff x=y =0$
– xbh
Nov 30, 2018 at 9:44
• @xbh: Oh, thanks! Didn't realize that! Nov 30, 2018 at 9:47
• For example $|2|+|-2| \ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$. Nov 30, 2018 at 12:33
• @GEdgar: Thanks for that! Nov 30, 2018 at 14:34

You dropped the absolute value bars between the second and third lines. If you have $$|f(t)| + \left| \frac{d}{dt} f(t) \right| = 0$$ for all $$t \in [a,b]$$, then it must be the case that both $$f(t) = 0$$ and $$\frac{d}{dt} f(t) = 0$$. Really, the first suffices, as this gives $$f(t) = 0$$ for all $$t \in [a,b]$$.