For $f \in L^1[0,1]$ let $A:L^1[0,1] \rightarrow L^1[0,1]$ be given by $$(Af)(x)=\int_0^x f(t)\,dt, \quad0\le x \le 1.$$ Show that $A$ is a continuous linear operator and calculate its norm.

I showed continuity, equivalently boundedness in this way: $$\|Af\|=\left\|\int_0^xf(t)\,dt\right\|=\int_0^1\left| \int_0^xf(t) \, dt \right| \,dx \le\int_0^1\int_0^x|f(t)| \, dt \, dx \le\int_0^1\|f\| \, dx=\|f\| $$ hence the operator is bounded with a constant $C=1$ and thus $\|A\|\le1$.

However, I'm struggling with showing that indeed $\|A\|=1$. I was advised to use $$f_n = n 1_{[0,{1 \over n}]}$$ but I have two questions. Firstly, how do I come up with such a function and what do I use it for?

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    $\begingroup$ Compute $\|f\|$ and $\|Af\|$ and compare their norm. How does one come up with this? These are standard examples to try. Specifically here they approximate the Dirac delta and their images under $A$ approximate a constant function. $\endgroup$ Jul 5, 2017 at 17:30
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    $\begingroup$ I have some qualms about your notation. You start with $\|(Af)(x)\|$. For each value of $x$, $(Af)(x)$ is simply a number. The object whose norm you seek is $Af,$ not $(Af)(x).$ So I'd have written $$\|Af\| = \int_0^1 |(Af)(x)| \, dx = \int_0^1 \left| \int_0^x f(t)\,dt \right| \,dx \le \int_0^1 \int_0^x |f(t)| \, dt \, dx = \cdots \text{ etc.}$$ $\endgroup$ Jul 5, 2017 at 17:59
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    $\begingroup$ What you use $f_n$ for is to show that $\dfrac{ \|Af_n\| }{\|f_n\|} \to 1$ as $n\to\infty,$ so you can conclude that $\|A\| \ge 1. \qquad$ $\endgroup$ Jul 5, 2017 at 18:04

1 Answer 1


It is clear that $(\forall n\in\mathbb{N}):\|f_n\|=1$. On the other hand, if $n\in\mathbb N$ and $x\in[0,1]$,$$(Af_n)(x)=\int_0^xf_n(t)\,dt=\left\{\begin{array}{l}nx&\text{ if }x\leqslant\frac1n\\1&\text{ otherwise.}\end{array}\right.$$Therefore, $\|Af_n\|=1-\frac n2$. So, you have a sequence of elements of your space with norm $1$ such that the norms of their images tendo to $1$. So, $\|A\|\geqslant1$. Since you already know that $\|A\|\leqslant1$, $\|A\|=1$.

  • $\begingroup$ This is a great "test function". So there is a natural question: If we restrict ourselves to $L^2[0,1]$, how to find $\|A\|$? I guess $\|A\|=1$ but such function doesn't work anymore. Can you give me some hints? $\endgroup$
    – Kimura Leo
    May 31, 2022 at 16:04
  • $\begingroup$ @Frankie Right now, I don't see an answer to that question. Perhaps that you could post it as a question. $\endgroup$ May 31, 2022 at 16:20

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