# Prove that $T$ is bounded and find its norm

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

Is it true?

Thanks a lot..

• Isn't $T:C([0,1])\to \mathbb{C}$ a bounded linear functional? – Song Dec 19 '18 at 8:57
• Actually, $\lVert T\rVert = \int_0^1 \sin tdt = 1-\cos(1)$ attained by $f=1$. – Song Dec 19 '18 at 9:13

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 :

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

• Thanks a lot 🌸 – Duaa Hamzeh Dec 19 '18 at 10:12

$$\|T\|$$ 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 $$\int \sin \, t dt$$

• Thanks a lot 🌸 – Duaa Hamzeh Dec 19 '18 at 10:12