Folland Real Analysis 5.2

Question

In Folland's second edition of Real Analysis, part of Proposition 5.2 (paraphrased) reads:

Let $\mathcal X$ and $\mathcal Y$ be normed vector spaces and $T:\mathcal X\to\mathcal Y$ a linear map. If $T$ is continuous at $0$, then $T$ is bounded.

His proof is as follows:

If $T$ is continuous at $0\in\mathcal X$, there is a neighborhood $U$ of $0$ such that $T(U)\subset\{y\in\mathcal Y:\|y\|\leq1\}$, and $U$ must contain a ball $B=\{x\in\mathcal X:\|x\|\leq\delta\}$ about $0$; thus $\|Tx\|\leq1$ when $\|x\|\leq\delta$. Since $T$ commutes with scalar multiplication, it follows that $\|Tx\|\leq a\delta^{-1}$ whenever $\|x\|\leq a$, that is, $\color{red}{\|Tx\|\leq\delta^{-1}\|x\|}$.

I do not understand how the highlighted part was obtained since $\|x\|\leq a$ (It would make sense to me if $\|x\|\geq a$.).

Answer 1

For the last conclusion, consider that $$|T(x)| \leq a\delta^{-1} \text{ whenever } |x| \leq a$$ $$\text{implies}$$ $$|T(x)| \leq |x| \delta^{-1} \text{ whenever } |x| \leq|x|$$ Since $|x| \leq |x|$ is always the case, Folland's logic is solid.

But I think more important is your intuition that the proof is somehow funky. The introduction of $a$ is not necessary at all.

When Folland reasons based on $T$ commuting with scalar multiplication, he's saying that any big vector is a scalar times a vector in the $\delta$ neighborhood of $0$. That is, write $$x = |x|\delta^{-1}x_\delta$$ where $x_\delta$ is the vector of length $\delta$ pointing in the same direction as $x$.

Then from what Folland already established using the definition of continuity at $0$, $$|T(x)| =T(|x|\delta^{-1}x_\delta) = |x|\delta^{-1}|T(x_\delta)| \leq |x|\delta^{-1}$$ This was all implicit in Folland's reasoning, and could be the end of the proof, but instead he obscures the proof. He trivially generalizes to $|x|\delta^{-1} \leq a \delta^{-1}$ whenever $|x| \leq a$, and then returns to the proper conclusion by specializing to the case |x| = a.

Answer 2

If $\|x\| \le a$ then $\|\frac{x \delta}{a} \| \le \delta$, so $\| T(\frac{x \delta}{a})\| \le 1 $ and $\|T(x)\| \le \frac a \delta$. In particular, apply this to $a = \|x\|$. (Which Folland should have done from the start - the argument is correct, but a bit too long.)

Answer 3

Here is one way of doing it:

Pick $x \neq 0$ and let $x' = {a \over \|x\|} x$, then $\|x'\| = a$.

Then $\|Tx'\| = \| Tx \| {a \over \|x\| } \le a {1 \over \delta}$, or $\|Tx\| \le {1 \over \delta} \|x\|$.

Answer 4

This may help:

If $\mathcal{X}$ and $\mathcal{Y}$ are normed vector spaces and $T:\mathcal{X}\rightarrow \mathcal{Y}$ is a linear map, TFAE:

a.) $T$ is continuous.

b.) $T$ is continuous at $0$.

c.) $T$ is bounded.

Proof - Suppose a.) holds. We can actually show that if we have contunity at a point $x_0$ then we have continuity everywhere. Take another point $x_1$. Given $\epsilon > 0$, take $\delta > 0$ such that $|x - x_0| < \delta$ implies $|Tx - Tx_0| < \epsilon$. Then for $|x' - x_1| < \delta$ we have $$|(x' + x_0 - x_1) - x_0| < \delta$$ by linearity if $T$, $$|Tx' - Tx_1| = |T(x' + x_0 - x_1) - Tx_0| < \epsilon$$ which is the desired contunity at $x_1$, so a.) implies b.).\ Now suppose b.) holds. For $\epsilon > 0$ there is a $\delta > 0$ such that $|x| < \delta$ implies $|Tx| < \epsilon$. For given $x\neq 0$, $$\left|\frac{\delta}{2|x|}x\right| < \delta$$ and so $$\left|T\frac{\delta}{2|x|}x\right| < \epsilon$$ thus by linearity of $T$ we have $$|Tx| < \frac{2\epsilon}{\delta}|x|$$ so $T$ is bounded. So b.) implies c.).\ Finally, suppose c.) holds. We will show that $T$ is then continuous at $0$. Suppose there is a $C\geq 0$ such that $|Tx|\leq C|x|$ for all $x$. Then, given $\epsilon > 0$, for $|x| < \epsilon/(C+1)$ $$|Tx|\leq C|x| < C\frac{\epsilon}{(C+1)} < \epsilon$$ thus $T$ is continuous at $0$.