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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) $ ?

1 Answers 1

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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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    As 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