Continuity of the functional on the space $L_1$.

Question

The norm of the space $\mathbb E = L_1(0,1)$ is $$\Vert f \Vert = \int_{0}^1 \vert f(t) \vert dt.$$ We have $$Tf(t) = \int_{0}^t f(t)dt. $$

Show that T is continuous.

I started from:

$\Vert Tf(t) \Vert = \int_{0}^1 \vert Tf(t) \vert dt = \int_{0}^1 \vert \int_{0}^t f(t) dt\vert dt \le \int_{0}^1 \int_{0}^t \vert f(t) \vert dt dt = \int_{0}^t \int_{0}^1 \vert f(t) \vert dt dt = \int_{0}^t \Vert f \Vert dt = \Vert f \Vert \int_{0}^t dt = \Vert f \Vert t. $

I get $ \Vert f \Vert t $ lastly. But I need to get $\Vert Tf(t) \Vert \le M \Vert f \Vert$, where $M>0$ (const). Can someone tell me where I made mistake?

Answer 1

Since $t\in[0,1]$, $\lvert f\rvert t\leqslant\lvert f\rvert$. So, take $M=1$.

Answer 2

A more elegant approach :

$$|Tf(t)| = \bigg|\int_0^1 f(t)\mathrm{d}t\bigg| \leq \int_0^1|f(t)|\mathrm{d}t \equiv \|f(t)\|$$

Thus, it is $\|Tf(t)\| \leq \|f(t)\| \implies \|T\| \leq 1$.

To show that $T$ is continuous, it suffices to show that $\|T\| = 1$.

Then, note that $\mathbf{1} \in S_{L_{1(0,1)}}$ and $T(\mathbf{1}) = I$ is the identity operator such that :$$1 = \|I\| = \|T(\mathbf{1})\| \leq \|T\|$$

But now, we have that : $$\| T\| \leq 1 \; \text{and} \; \|T \| \geq 1 \implies \|T\| =1 \implies T \; \text{continuous}$$