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Suppose that $X, Y$ are normed linear space and that $T:X\to Y$ is a linear map, then prove that

a) $T$ is continuous on $X$ iff $T$ is continuous at $0$.

and using a), prove that $T$ is continuous on $X$ iff there exists a $C>0$ such that $||Tx||\le C||x||$ for all $x\in X$.

I tried to prove it as follows:

a) ($\Rightarrow$) is straightforward.

($\Leftarrow$) Let $x\in X$ be arbitrary. Let $x_n\to x$. Suppose that $r_n:=x_n-x$. Clearly, $r_n\to 0$. $Tr_n+Tx=Tx_n$. By continuity of $T$ at $0$, $Tr_n\to 0$; and therefore, it follows that $Tx_n\to Tx$. Since $x$ is arbitrary, it follows that $T$ is continuous on $X$. This completes the proof of part $(a)$.

The second part is where I get stuck.

($\Leftarrow$) This direction is straightforward.

$(\Rightarrow)$ By a), $T$ is continuous at $0$ so there is a $\delta>0$ such that $||Tx||\lt 1$ for all $x: ||x||<\delta$. I don't know how to proceed from here to introduce $C$. Any hints on this? Thanks.

Koro
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3 Answers3

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As you correctly pointed out, there exists a $\delta>0$, s.t. if $||x||<\delta$ we have $||Tx||<1$.

Let $y\in X$, with $y\ne 0$. Now if we define $x:=\frac{\delta}{2}\frac{y}{||y||}$ then of course $||Tx||<1$, since $||x||<\delta$.

However: $||Ty||=||\frac{2||y||}{\delta}Tx||$ and from that on I'm sure you'll get it yourself :)

Edit: Sorry haven't seen that is has already been answered.

SetofMeasureZero
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  • That's a fantastic answer +1. Thank you very much. In the second line however, $||x||=\delta$ but that can be fixed by choosing $\frac {\delta}2$ instead of $\delta$. :-) – Koro Jun 15 '22 at 16:55
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    Oh you're absolutely right. I'll fix that right away. – SetofMeasureZero Jun 15 '22 at 16:57
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Choose $x_0\in X$ .

$\|Tx-Tx_0\|=\|T(x-x_0)\|$

Since $T$ is continuous at $0$, $\|(x-x_0)\|<\delta$ implies $\|T(x-x_0)\|<\epsilon$

Suppose $\forall C>0, \exists x\in X$ such that $\|Tx\|\ge C\|x\|$

Hence $\forall n\in\Bbb{N}, \exists( x_n) \subset X$ such that $\|Tx_n\|\ge n\|x_n\|$

Then $y_n=\frac{x_n}{n\|x_n\|}\to 0$ but $\|Ty_n\| $ doesn't converge to $0$ which contradict the continuity of $T$ at $0$.

Infact if $T$ is a bounded linear map $\|Tx\|\le \|T\|_{op}\|x\|$

Where $\|T\|_{op}=\sup\{\|Tx\|: \|x\|\le 1\}$

Sourav Ghosh
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  • Thanks for the answer :).Can you please explain how that gives me $C$ that works for all $x\in X$? – Koro Jun 15 '22 at 16:48
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Contrary to your claim, suppose that for every positive integer $n$ there exists $x_n \in X$ such that $\|Tx_n\| > n\|x_n\|$. Since $T$ is linear, we may assume that $\|x_n\| = 1$ for every $n$ (if not, just take $x_n' = x_n/\|x_n\|$). Hence, $$ \|T \frac{x_n}{\sqrt{n}}\| > \sqrt{n} $$ for every $n$, which contradicts the continuity of $T$ at $0$.

xen
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