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I did the following exercise (given in my notes) can you tell me if my answer is correct? Thanks.

Exercise: Compute the operator norm of the continuous map $f \mapsto f$ when viewed:

(a) as a map $T: C^1([0, 1]) \to C([0, 1])$

(b) as a map $S: C([0, 1]) \to L^1_\mu([0, 1])$, where $\mu$ is Lebesgue measure on $[0, 1]$

(c) Compute the operator norm of the composition of the maps from (a) and from (b).

(d) Now restrict the maps in (a), (b) and (c) to the space of functions $f$ with $f(0) = 0$, and compute the operator norms again.

My answer:

(a) $\|T\| = \sup_{\|f\|_{C^1} \leq 1} \|f\|_\infty \leq \sup_{\|f\|_\infty \leq 1} \|f\|_\infty = 1$ since for example $f(x) = x \in C^1[0,1]$ and $\|f\|_\infty = 1$. OTOH, $\|T\| \geq 1$ since $\|Tf\| = \|x\| = 1$.

(b) $\|S\| = \sup_{\|f\|_\infty \leq 1} \|f\|_1 = 1$ since $\|f\|_1 \leq \int_{[0,1]} 1 d \mu$ so $\sup_{\|f\|_\infty} \|f\|_1 \leq 1$. But $f(x) = 1 \in C[0,1]$ and $\|1\|_\infty \leq 1$ and $\|1\|_1 = 1$.

(c) $\|ST\| \leq \|S\| \|T\| = 1$. We have $\|ST\| \geq 1$ since for $f(x) = 1$, $\| STf\| = \|Sf\| = 1$.


(d.a) $\|T\| = 1$ by the same argument as in (a).

(d.b) $\|S\| = 1$ because the sequence $f_n$ where $f_n$ is the function that is $nx$ on $[0, \frac1n]$ and $1$ on $[\frac1n, 1]$ is in $C[0,1]$ and $\|f_n\|_\infty \leq 1$ and $\|f_n\|_1 \to 1$ (by monotone convergence theorem)

(d.c) $\|ST\| \leq 1$ (same argument as (c)). Unfortunately, $f_n$ from (d.b) are not in $C^1[0,1]$. So I'm not sure what to do. Is $\|ST\| = 1$ in this case too? Or perhaps not?

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Dear Matt N. shouldn't we consider $C^1([0,1])$ with the norm $||f||=||f||_{\infty}+||f'||_{\infty}$? – Giuseppe Jul 15 '12 at 10:17
Hm. The exercise doesn't say. But you are probably right. Let me redo the exercise with the other norm. @GiuseppeTortorella and t.b. – Rudy the Reindeer Jul 15 '12 at 10:19
No, it's fine, I confused myself. I'd prefer to write it in two steps $$\lVert f \rVert_1 = \int |f|\,d\mu \leq \int \|f\|_\infty\,d\mu = \lVert f \rVert_\infty.$$ – t.b. Jul 15 '12 at 10:23
Ok. : ) ${}{}{}{}{}$ – Rudy the Reindeer Jul 15 '12 at 10:25
The argument stays more or less the same. But perhaps I am missing something? – Rudy the Reindeer Jul 15 '12 at 11:27
up vote 0 down vote accepted

(Using the norm $\|f\|_{C^1}=\|f\|_\infty+\|f'\|_\infty$

(d.a) by mean value theorem, we have $\|f\|_\infty\le\|f'\|_\infty$, since for all $x$, $|f(x)-f(0)\\le\|f'\|_\infty|x-0|\le\|f'\|_\infty$, so $2\|f\|_\infty\le\|f\|_\infty+\|f'\|_\infty=1$, and hence, $ \|T\|\le1/2$. But $f$, defined by $f(x)=\frac{x}{2}$ satisfy $\|f\|_{C^1}=1$ and $\|f\|_\infty=1/2$, then $\|T\|=1/2$.

(d.c) let $f\in C^1$ with norm $1$. By previous argument, $\|f\|_\infty\le1/2$. If for any $y$ we have $f(y)>y/2$, then there exist $c$ such that $f'(c)=f(y)/y$, so if $\|f\|_\infty=1/2$, then $\|f\|_{C^2}>1$. From this, we have $\|ST\|=1/4$.

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