# Norm of integral operator in $L^1$

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

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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 $||T||=1$.

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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