# Why $\vert T(x)\vert \leq \left\Vert T \right\Vert \left\Vert x \right\Vert$?

$\vert T(x)\vert \leq \left\Vert T \right\Vert \left\Vert x \right\Vert$ is a frequently-used inequality. Here, $x\in X$, $X$ is a normed space. $T$ is a functional. But I don't know how to relate $\left\Vert \cdot \right\Vert$ with $\left\vert \cdot \right\vert$. I know that $$\left\Vert T \right\Vert= \sup_{\left\Vert x \right\Vert=1} \left\Vert T(x) \right\Vert = \sup_{\left\Vert x \right\Vert \neq0}\frac{\left\Vert T(x) \right\Vert}{\left\Vert x \right\Vert}$$ Next, then I wonder why $\left\Vert T(x) \right\Vert \leq \left\Vert T \right\Vert \left\Vert x \right\Vert$, which I use frequently as well. However, I can't see how to prove it from the definition of norm space, and it looks like Hölder's inequality without specifying the norm. Can some give me some help? Thanks a lot.

Edit:

About the inequality $\vert T(x)\vert \leq \left\Vert T \right\Vert \left\Vert x \right\Vert$, is it because that: if $0<p<q<\infty$ and $E$ is measurable, then $L^q(E) \subset L^p(E)$, therefore, $L^q(E) \subset L^1(E)$. But it is true only for $L^p$-norm.

• What does $\sup$ mean?
– user228113
May 12, 2018 at 15:39
• That wasn't him being ignorant, that was him helping you.
– user223391
May 12, 2018 at 15:41
• @G.Sassatelli is giving you a hint to understanding the inequality. May 12, 2018 at 15:41
• The norm on $T$ is defined in such a way that this inequality has to be true. If there were some $x$ not satisfying it, then the value we picked for the $\sup$ wasn't really a $\sup$. May 12, 2018 at 15:44
• Oh, indeed. The second inequality is directly from moving the scalar ||x||, thanks May 12, 2018 at 15:46

The case where $x=0$ is trivial.
Observe that for $x\not=0$, $$\|T(x)\|=\left\|\|x\|T\left(\frac{x}{\|x\|}\right)\right\|=\|x\|\left\|T\left(\frac{x}{\|x\|}\right)\right\|\leq\|x\|\sup_{\|y\|=1}\|T(y)\|=\|x\|\|T\|.$$