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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..

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  • $\begingroup$ Isn't $T:C([0,1])\to \mathbb{C}$ a bounded linear functional? $\endgroup$ – Song Dec 19 '18 at 8:57
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    $\begingroup$ Actually, $\lVert T\rVert = \int_0^1 \sin tdt = 1-\cos(1)$ attained by $f=1$. $\endgroup$ – Song Dec 19 '18 at 9:13
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Vastly updated answer, providing a more elementary approach than the rather completely correct and elegant approach of Kavi Rama Murthy :

The linearity is easy to show (which you haven't done). Take $f, g \in C[0,1]$ and $\lambda \in \mathbb R$ and prove that $T(\lambda f+g) = \lambda Tf + Tg. $

For the boundedness, note that it is :

$$|Tx| = \bigg|\int_0^1 \sin(t)f(t) \mathrm{d}t \bigg| \leq \int_0^1 |\sin(t)f(t)|\mathrm{d}t \leq \|f\|_\infty\int_0^1 |\sin(t)|\mathrm{d}t $$

But, note that $\sin(t) \geq 0$ for $t \in [0,1]$, thus it is :

$$\|Tx\| \leq \|f\|_\infty\int_0^1\sin(t)\mathrm{d}t$$

That means that $T$ is a bounded linear operator $T : C[0,1] \to \mathbb R$ with $\|T\| \leq \int_0^1 \sin(t)\mathrm{d}t$.

Now, take $\mathbf{1} \in C[0,1]$. Then, it is :

$$T(\mathbf{1}) = \int_0^1\sin(t)\mathrm{d}t \implies \|T(\mathbf{1})\| = \int_0^1\sin(t)\mathrm{d}t \implies \|T\| = \int_0^1 \sin(t)\mathrm{d}t$$

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  • $\begingroup$ Thanks a lot 🌸 $\endgroup$ – Duaa Hamzeh Dec 19 '18 at 10:12
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$\|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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  • $\begingroup$ Thanks a lot 🌸 $\endgroup$ – Duaa Hamzeh Dec 19 '18 at 10:12

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