# Equivalent norms in $~C^1([0,1])~$

I have to prove that the following norms are equivalent in $$C^1([0,1])$$: $$|| f ||_* = |f(0)| + \max_{t \in[0,1]} |f'|\\ || f||_\infty = \max_{t \in[0,1]} |f|$$

I have: \begin{align} || f||_\infty &= \max_{t \in[0,1]} |f| \\&= \max_{t \in[0,1]} \left|f(0) + \int_{0}^{t}f'(\tau)d\tau\right| \\ &\le |f(0)| + \int_{0}^{t}|f'(\tau)|d\tau \\&\le |f(0)| + \int_{0}^{1} |f'(\tau)|d\tau \\&\le |f(0)| + \max_{t \in [0,1]}|f'(t)| \\&= ||f||_*\end{align} Now I'm in trouble to find the other inequality.

Another little question. I know that $$||f||_\infty$$ and $$||f||_1$$ are not equivalent in $$C([0,1])$$ but are they equivalent in $$C^1([0,1])?$$ I'm asking since on some notes I'm asked to prove that the $$||f||_*$$ is equivalent to both these two norms, but it may be an error.

• These norms do not seem to be equivalent. Consider $f_n(x)=\frac{\sin{n^2x}}{n}$, then $\|f_n\|_{\infty}=1/n$, while $\|f_n\|_*=n$. – Mindlack Aug 26 '19 at 8:35

You cannot prove it, since it is false. If $$f_n(x)=\sin(nx)$$, then $$\lVert f_n\rVert_\infty\leqslant1$$, whereas $$\lVert f_n\rVert_*=n$$.
The same sequence of functions can be used to prove that the norms $$\lVert\cdot\rVert_\infty$$ and $$\lVert\cdot\rVert_1$$ are not equivalent in $$\mathcal C^1\bigl([0,1]\bigr)$$.
• Nothing is the OP's question is meaningless, as far as I understand what he wrote. There is no mention of a norm on $\mathcal{C}([0,1])$ using the derivative. May be the last sentence of his post is a little confusing I agree. – nicomezi Aug 26 '19 at 8:47
• Thanks. Yes, yuo're right. Actually the text of the exercise is: prove that $||f||_*$ is equivalent to the "usual norm" in C^1([0,1]), I presumed it was the infinite norm. In the other notes I figured out that it isn't asking for both 1-norm and infinite norm but for $||f||_{1,\infty} = ||f||_\infty + ||f'||_\infty$ which is not hard. Thanks, I'm sorry for the mistake. – Peanojr Aug 26 '19 at 9:09
• In the last sentence I was asking if those two norms are equivalent in $C^1([0,1])$ since if they are, then it's enough to prove that the * norm is equivalent to only one of those and then use transitivity for the other. – Peanojr Aug 26 '19 at 9:12