Prove that $T$ is bounded and find its norm

functional-analysisnormed-spacesoperator-theory

Consider $T\colon C[0,1] \to C[0,1]$ by

$$T(f) = \int_{0}^1 \sin(t) \;f(t) \;dt$$

  • Show that $T$ is a bounded linear operator.
  • Find the norm of $T$.

This is my solution

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Is it true?

Thanks a lot..

Best Answer

$\|T\| $ is not $1$. In fact $\sin \, t$ is positive and increasing in $(0,1)$ so $|Tf| \leq \sin (1) \|f\|$. Hence $\|T\| \leq \sin (1)$. Actually, $\|T\|=\int_0^{1}\sin\, t \, dt$. To prove this you have to use the fact that $(L^{1} [0,1])^{*}=L^{\infty}([0,1])$ and the fact that functions in $L^{\infty}([0,1])$ can be approximated in $L^{1}([0,1])$ norm by continuous functions whose sup norms don't exceed the norm of the original function.

Here is a detailed argument: there exist a sequence $\{f_n\}$ in $ L^{\infty}([0,1])$ such that $\int f_n(t)\sin \, t dt \to \int_0^{1}\sin\, t \, dt$ and $\|f_n\|_{\infty} \leq 1$ for all $n$. Since $f_n$'s are also in $L^{1}([0,1])$ There exist continuous functions $g_n$ such that $\|g_n\|_{\infty} \leq 1$ and $\int|f_n-g_n| \to 0$. Hence $\lim \inf \int_0^{1} g_n(t) \sin \, \, dt \geq \lim \inf \int_0^{1} f_n(t) \sin \,t \, dt =\int\sin \, t \, dt$.

PS After seeing Rebellos answer I have realized that I am making things too complicated. Just taking $f=1$ will show that the norm is $\int \sin \, t dt$

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