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For $f \in L^1[0,1]$ let $$(Af)(x)=\int_0^xf(t)\,dt, \quad0\le x \le 1.$$ Show that in both following cases $A$ is a continuous linear operator between the given spaces and calculate its norm.

a) $\quad A:L^1[0,1] \rightarrow L^1[0,1];$

b) $\quad A:C[0,1] \rightarrow C[0,1]$.

My attempts at showing continuity, equivalently boudnedness:

a)

$||(Af)(x)||=||\int_0^xf(t)\,dt||=\int_0^1| \int_0^xf(t) \, dt | \,dx \le\int_0^1\int_0^x|f(t)| \, dt \, dx \le\int_0^1||f|| \, dx=||f|| $, hence the operator is bounded with a constant $C=1$.

b)

$ ||(Af)(x)||=||\int_0^x f(t) \, dt||=\sup \limits_{x\in[0,1]} |\int_0^x f(t) \, dt| \le \sup \limits_{x \in [0,1] } \int_0^x |f(t)| \, dt=\int_0^1 |f(t)| \, dt \le (1-0) \cdot \sup \limits_{t\in[0,1]} |f(t)| \, dt=||f||,$ hence the operator is also bounded with $C=1$.

My question is if my work so far has been correct and if you could give me a hint how to calculate the norms as I get a little lost using $||A||=\sup \limits_{||f||=1} ||(Af)(x)||$.

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1 Answer 1

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Generally you are looking for unit norm $f$ such that $\|Af\| = C$ or else for any $\epsilon >0 $, some unit norm $f$ such that $\|Af\| > C - \epsilon$.

For a) try to find a non negative $f_n \in L^{1}[0,1]$ of norm 1 whose support is contained in a neighbourhood of $0$ (think 'delta functions').

Spoiler:

Try $f_n = n 1_{[0,{1 \over n}]}$.

For b) try to find a non negative unit norm $f \in C[0,1]$ such that $\int_0^1 f = C$.

Spoiler:

Try $f = 1$.

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  • $\begingroup$ I'm afraid I don't get it, could you explain what definition you make use of and what are the $C$'s you used? $\endgroup$
    – Theta
    Jul 3, 2017 at 16:15
  • $\begingroup$ @Theta: I am using the same definitions & $C$s (both $1$) that you used in your question. $\endgroup$
    – copper.hat
    Jul 3, 2017 at 19:39
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    $\begingroup$ The definition is $\|A\| = \sup_{\|f\| \le 1} \|Af\|$ where the appropriate norm is used. You showed that $\|A\| \le C = 1$. To show equality you need to show that $\|A\| \ge C-\epsilon$ for all $\epsilon>0$. $\endgroup$
    – copper.hat
    Jul 3, 2017 at 20:05
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    $\begingroup$ You want to maximise $\int_0^1 |\int_0^x f(t)dt|dx$ subject to $\int_0^1 |f(x)| dx = 1$. Since the inner integral 'ignores' values of $f$ for $t < x$ we would like (loosely speaking) for $f$ to have a 'large' area near 0. Pick $f$ to be non negative to see if that will work. $\endgroup$
    – copper.hat
    Jul 5, 2017 at 19:44
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    $\begingroup$ Yes, I meant $t>x$, thanks! $\endgroup$
    – copper.hat
    Jul 5, 2017 at 21:11

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