# Show that $\infty$-norm and $C^1$-norm are not equivalent.

Show that $\infty$-norm and $C^1$-norm are not equivalent.

For the $C^1([a,b],\mathbb{R})$ space, show that

$\displaystyle ||g||_\infty=sup_{a\leq t\leq b}|g(t)|$

and

$\displaystyle ||g||_{C^1}=sup_{a\leq t\leq b}|g(t)|+sup_{a\leq t\leq b}|g'(t)|$

are not equivalent.

My attempt:

$\displaystyle |g'(t)|>0 \implies sup_{a\leq t\leq b}|g'(t)|>0$

So then

$\displaystyle ||g||_\infty=sup_{a\leq t\leq b}|g(t)| \leq sup_{a\leq t\leq b}|g(t)|+sup_{a\leq t\leq b}|g'(t)| \leq p ||g||_{C^1}$ for constant $p \geq 1$.

If the two norms are not equivalent then I'm assuming that

$||g||_\infty \ngeq q ||g||_{C^1}$ for any constant $0<q<1$. How do I show this is true?

Is this a good approach or does anyone have a better idea?

• Does $g(x)=x^{\frac{1}{3}}$ work? g'(x)= is unbounded but it's not continuous at x=0 though? Sep 23 '14 at 14:02
• What you need is not just one function, but a collection! To prompt you a little more, what are the extreme values of $f_n(x) = sin(nx)$? What are the extreme values of their derivatives? Sep 23 '14 at 14:03
• Ok so your $f_n \in [-1,1]$ and $f_n'(x)=ncos(nx) \in [-n,n]$. The derivative is unbounded as $n$ tends to $\infty$. Sep 23 '14 at 14:17
• Then $||fn||_\infty = sup|f_n(x)|=1$ and $||fn||_{C^1} = sup|f_n(x)|+sup|f_n'(x)| \to \infty$. Sep 23 '14 at 14:20
• Then $||f_n||_\infty \geq q||f_n||_{C^1}$ cannot be true for any $q>0$ ? And thus, there is no $q>0$ that works for all functions in $C^1$ ? Sep 23 '14 at 14:26