= Solution
Both maxima exist by continuity on the <compact set> $[a,b]$. Their sum is nonnegative and vanishes only for $f=0$. The identities $\|cf\|_1=|c|\|f\|_1$ and the <triangle inequality> follow by applying the corresponding <supremum norm> facts to $f$ and $f'$. This proves that $\|\cdot\|_1$ is a <norm>.
For $m=(a+b)/2$, $|\Phi(f)|=|f'(m)|\leq\max|f'|\leq\|f\|_1$. Hence \b[the first unit ball has image contained in $[-1,1]$]: <derivative> evaluation is a <bounded linear functional> for this <norm>.
For the <supremum norm> alone, take $f_N(x)=\sin(N(x-m))$. These smooth functions satisfy $\|f_N\|_0\leq1$, but $\Phi(f_N)=N$. Thus \b[the second image is unbounded]. The distinction is the <unboundedness of derivative evaluation in the supremum norm>, not a failure of differentiability of the individual functions.
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