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Let be $E$ a real Banach space and $E^*$ the real dual space. $\forall T \in E^*$

Is it true $\left \| T(x) \right \| \le \left \| T \right \| \cdot \left \| x \right \| $? Why?

Let be $\{ T_n \}_n$ a Cauchy's sequence. Is it true $\left \| T_n(x)-T_m(x) \right \| \le \left \| T_n-T_m \right \| \cdot \left \| x \right \| $? Why?

Thanks to all.

NOTE: In my book (Brezis' one) the norm is define as: $$ \left \| T \right \|= \sup_{\left \| x \right \| \le 1} \left \| T(x) \right \| $$

Skills
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    What? I know that $\left | T \right |= \sup_{\left | x \right | < 1} \left | T(x) \right |$ – Skills Dec 13 '15 at 17:55
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    Hint for the first part: because $T$ is linear, $T(ax)=aT(x)$ for scalar $a$. So now replace your (nonzero) $x$ with $| x | \frac{x}{| x |}$. – Ian Dec 13 '15 at 18:04

2 Answers2

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Yes. For $x \neq 0$:

$$ \| T(x) \| = \left \| T \left(\frac{x \|x\|}{\|x\|} \right) \right \| = \| x \| \left \|T \left(\frac{x}{\|x\|} \right) \right \| \le \|T\| \|x\|$$

And it also obviously holds for $x=0$

Ian
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S -
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  • Is the second true because $(T_1+T_2)(x)=T_1(x)+T_2(x)$? – Skills Dec 13 '15 at 18:16
  • @Skills The second equality is because of the linearity of $T$. If $\lambda \in \mathbb{K}$ then $T(\lambda x)= \lambda T(x)$. In this case $\lambda = | x |$ – S - Dec 13 '15 at 18:17
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    @Skills Ah, if you refer to your second question it is indeed true for the reason that you mention. – S - Dec 13 '15 at 18:24
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In general, we can write this: $$\|T\| = \sup_{z \neq 0} \frac{\|T(z)\|}{\|z\|}.$$ Suppose that $z^*$ is the value of $z$ for which you have the sup. Obvioulsy:

$$\|T\| = \frac{\|T(z^*)\|}{\|z^*\|}.$$

Consider a generic $x$. Then:

$$\frac{\|T(x)\|}{\|x\|} \leq \frac{\|T(z^*)\|}{\|z^*\|} = \|T\| \Rightarrow \|T(x)\|\leq\|T\| \|x\|.$$

Notice that using different definition as: $$\|T\| = \sup_{\|z\| \leq 1} \frac{\|T(z)\|}{\|z\|},$$

doesn't change the situation. Indeed, suppose that $\|z^*\| = M > 1$. Then consider a vector $y^* = \frac{1}{M} z^*$, so that $\|y^*\| = 1$. In this case:

$$\frac{\|T(y^*)\|}{\|y^*\|} = \frac{\|T\left(\frac{1}{M}z^*\right)\|}{\|\frac{1}{M}z^*\|} = \frac{\frac{1}{M}\|T\left(z^*\right)\|}{\frac{1}{M}\|z^*\|} = \frac{\|T(z^*)\|}{\|z^*\|}.$$

This means that also $y^*$ is a point at which the sup is attained.

In general, the definition of sup is:

$$\|T\| = \sup_{\|z\| =1} \frac{\|T(z)\|}{\|z\|}.$$

Your book is giving an equivalent formulation. That's all.

Let's talk about the sequence. Since all $T_n$ are linear transformations, then $T_n - T_m$ is a linear transformation too (i.e. $T_n - T_m \in E^*$).

Using the previous results, you have that:

$$\|T_n(x) - T_m(x)\| \leq \|T_n - T_m\| \|x\|.$$

the_candyman
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