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I am studying a theorem about linear maps on normed vector spaces. The theorem states the equivalence of the following statements:

$T$ is a linear map on a normed vector space $X$ that maps to a normed vector space $Y$.

  1. $T$ is continuous
  2. $T$ is continuous at zero.
  3. $T$ is bounded.

Clearly, 1 $\implies$ 2 is immediate. And 3 $\implies$ 1 is not so complicated either: suppose $\|Tx\| \le C\|x\|$, so $\|Tx-Ty\| = \|T(x-y)\| \le C\|x-y\|$, so take $\delta = \frac{\epsilon}{C}$, and $\|x-y\| < \delta \implies \|Tx-Ty\| < \epsilon$.

However, I am having some difficulty following a proof of 2 $\implies$ 3. The author makes the following claim:

If $T$ is continuous at $0 \in X$, there is a neighborhood $U$ of $0$ such that $T(U) \subset \{y \in Y : \|y\| = 1\}$, and $U$ must contain a ball $B = \{x \in X : \|x\| \le \delta\}$ about 0; thus $\|Tx\| \le 1$ when $\|x\|\le \delta$.

This is clear; it is just a restatement of the Weierstass criterion for continuity. He proceeds:

Since $T$ commutes with scalar multiplication, it follows that $\|Tx\| \le a\delta^{-1}$ whenever $\|x\| \le a$, that is, $\|Tx\| \le \delta^{-1}\|x\|$.

This I do not follow. What is happening here? Is this just a rescaling $x \mapsto \frac{a}{\delta} x$? I cannot replicate this conclusion.

user0
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Emily
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1 Answers1

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Suppose $\|x \| = a$; write $x = \frac{a}{\delta} y$, where $\|y\| = \delta$. You know that $T(x) = \frac{a}{\delta} T(y)$, and $\| T(y) \| < 1$. So $\|T(x)\|< |\frac{a}{\delta}|$.

The case where $\| x \| < a$ is similar.

John Hughes
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