Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the norm of the operator $$ T\colon L^1[0,1] \to L^1[0,1]: f\mapsto \left(t\mapsto \int_0^t f(s)ds\right) $$ ?

share|cite|improve this question
up vote 6 down vote accepted

Let $f\in L^1([0,1])$. Then

$$\|Tf\|_1=\int_0^1 \left|\int^t_0 f(s) ds\right| dt \le \int_0^1 \int_0^1 |f(s)| ds dt = \|f\|_1$$

This shows $\|T\|\le 1$. Setting $f_n(x)=n\chi_{[0,1/n]}(x)$, we see $||f_n||_1=1$. Note that

$$\int^t_0 n\chi_{[0,1/n]}(s) ds=\left\{\begin{array}\,1 & \text{if}\;t\ge1/n\\ nt & \text{if}\;t<1/n\end{array}\right.$$ It follows that $$||Tf_n||_1=\int^1_0\int_0^t n\chi_{[0,1/n]}(s)ds dt=\int_0^{1/n}nt\,dt+\int_{1/n}^1 1\,dt =1-\frac{1}{2n}\rightarrow 1\;\text{as}\;n\rightarrow\infty. $$ Hence $||T||=1$.

share|cite|improve this answer
For $f \equiv 1$, I found $||Tf||_1=1/2$ and not $1$. Otherwise, $f(x)=e^x$ works. – Seirios Oct 15 '12 at 16:46
Thanks for the answers! But by the definition of the norm of T as $\Vert T \Vert = \sup_ {f \in L^1[0,1],\Vert f\Vert_1=1} \Vert Tf \Vert_1$, I can't really see how $f=2$ works, nor can I see that using $f=e^x$ will work, since $\Vert f \Vert _1 \neq 1$ for both of these cases. Is the definition I'm using wrong? – Maethor Oct 15 '12 at 20:17
What makes you think that your simple-minded estimate is anywhere near sharp? If you do just one step in your computation, you get $$ \int_{0}^{1}\left\lvert\int_{0}^t f(x)\,dt\right\rvert\,dx \leq \int_{0}^1 \int_{0}^t \lvert f(x)\rvert\,dx\,dt $$ which is an integral over a triangular region. If you brutally replace $t$ by $1$ here, you will get an integral over a square, so your estimate will overshoot badly. I would suggest to think about $F(t) = \int_{0}^t f(x)\,dx$ and think about what differential equation it solves. You should get $2/\pi$ as a final solution, unless I'm mistaken. – commenter Oct 16 '12 at 2:27
@Norbert: geeez, you're right... I was led astray by considering $f(x) = \frac{\pi}{2} \cos{\frac{\pi x}{2}}$ as in the $L^2$-case. .@IHaveAStupidQuestion: Try $f_n = n \chi_{[0,1/n]}$. This sequence will minimize the effect of charging the upper right triangle in your estimate and a computation shows that $\lVert Tf_n\rVert_{1} \nearrow 1$. – commenter Oct 16 '12 at 11:40
@MattN. The operator $T$ is the Volterra operator. The trick to compute its norm in $L^2$ is to consider $S = T^\ast T$. Then $\lVert T\rVert^2 = \lVert T^\ast T\rVert$. Use that $S$ is compact and self-adjoint, so its norm is equal to its maximal eigenvalue. An eigenfunction $\lambda f = T^\ast T f$ is a solution to $f'' = - \lambda f$ and this yields an ansatz that lets you compute the eigenvalues and eigenvectors of $S$ and thus its norm. – commenter Oct 16 '12 at 13:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.