What is the norm of the operator $$ T\colon L^1[0,1] \to L^1[0,1]: f\mapsto \left(t\mapsto \int_0^t f(s)ds\right) $$ ?
Norm of integral operator in $L^1$
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functional-analysis
operator-theory
norm
1 Answers
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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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1For $f \equiv 1$, I found $||Tf||_1=1/2$ and not $1$. Otherwise, $f(x)=e^x$ works. – 2012-10-15
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0@Seirios I think $f= 2$ works also. – 2012-10-15
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1Thanks 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? – 2012-10-15
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0@Maethor Let $f=2$. Then $\|Tf\|_1 = \int_0^1 \int_0^t 2 dx dt= 2 \int_0^1 t dt = 2\frac{t^2}{2} \mid_0^1 = 1$. Unfortunately, this does indeed not help here since $\|f\|_1 = 2$. – 2012-10-15
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0But the idea is that you find an $f$ such that both $\|Tf\|_1$ and $\|f\|_1$ are $1$. – 2012-10-15
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0@Matt N. No worries :) Yeah, I've been trying to do find an $f$ such that $\Vert Tf \Vert _1 = \Vert f \Vert _1 = 1$ for some time, but with no luck. Any further help is much appreciated. – 2012-10-15
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0@Maethor Maybe something with $\sin $? Have you tried that already? – 2012-10-15
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0Aha. You got a down vote. (not by me) That's probably to tell you that your answer wants fixing and the down voter will probably un-down vote once you correct it. – 2012-10-15
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0Actually, isn't there a typo in your inequality? You want a $t$ instead of a $1$ in the boundaries of your integral. – 2012-10-15
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0@Matt N. I'm sorry, what? What should I correct? My answer? You mean the answer provided by IHaveAStupidQuestion? Sorry, I'm new to the whole stackExchange thing. – 2012-10-15
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0@Maethor Sorry for the confusion: yes, I meant IHaveAStupidQuestion's answer and was talking to him in my last two comments. – 2012-10-15
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1What 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. – 2012-10-16
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0@commenter you will het $2/\pi$ for the operator on $L_2[0,1]$ – 2012-10-16
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0@Norbert I'm sorry but do you know what commenter means by what differential equation? I know almost nothing about DEs and I think I'm stuck. – 2012-10-16
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3@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 $\lVert Tf_n\rVert_{1} \nearrow 1$. – 2012-10-16
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1@MattN. I see the problem resolved – 2012-10-16
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0@commenter: Thanks a lot! I included it above. – 2012-10-16
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0Dear @Norbert thanks for the ping. Just to be clear: was commenters comment wrong? Or what is the differential equation? I still don't see : / – 2012-10-16
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3@MattN. The operator $T$ is the [Volterra operator](http://math.stackexchange.com/a/151444/). 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. – 2012-10-16
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0To answer my own question: yes, commenter's previous comment was misleading since we do not need DEs to compute $\|T\|$ if $T: L^1 \to L^1$. – 2012-10-18
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0As for your last comment: I'll ignore that for now, it's tmi. I don't understand how you see from $\lambda f = T^\ast T f$ that an eigenfunction is one such that $f'' = - \lambda f$. (Unless $T^\ast = T$ but I don't know how to compute complex conjugate of an operator) – 2012-10-18