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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.

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

1 Answer 1

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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 $t \in [a,b]$.

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    $\begingroup$ Thanks, platty! $\endgroup$ Nov 30, 2018 at 9:48

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