I want to prove the following statement.

Let $T\colon X\to Y$ be a linear operator between normed spaces, and suppose that $T$ is sequentially continuous, i.e., if $(x_n)$ is a sequence in $X$ with $x_n\to x$, then $Tx_n\to Tx$.

Then $T$ is bounded, i.e., there exists $c>0$ such that for all $x\in X$, $$\|Tx\|\leqslant c\,\|x\|.$$

The following is my attempt.

Let $x\in X$, and consider the sequence $(\frac xn)_{n\in\mathbb N}$. Then $x_n\to0$, so $Tx_n\to T0=0$. Consequently, there exists $N$ such that $\|Tx_n\|<\|x\|$ for all $n\geqslant N$. In particular, $$\|Tx_N\| = \|T\tfrac{x}{N}\| = \frac1N\|Tx\| < \|x\| \implies \|Tx\|<N\|x\|,$$ so $T$ is bounded.

Is this argument valid? If so, is there a shorter argument to prove this?


Your $c$ depends on $x$, which can't happen (look at the order of the quantifiers). Try it by contradiction, instead.

Suppose that, for any $c>0$, there exists $x$ so that $\|Tx\|>c\|x\|.$ We claim that there exists a convergent sequence $(x_n)$ such that $\|Tx_n\|\rightarrow\infty.$ Indeed, if $c=1,$ there there exists $x_1$ with norm $1$, so that $\|Tx_1\|>\|x_1\|=1.$ If $c=4,$ then there exists $x_2$ with norm $1$ so that $\|Tx_2\|>4\|x_2\|=4.$ Proceeding inductively generates a sequence of elements $(x_n)$ with norm one so that $\|T x_n\|>n^2.$ Now, take $z_n=\frac{1}{n}x_n.$ Note that $z_n$ converges to zero, and $\|Tz_n\|>n.$ That is, we have found a sequence $z_n\rightarrow 0$, but $Tz_n\not\rightarrow T(0)=0.$

  • $\begingroup$ How exactly is the sequence $x_n$ defined? $\endgroup$ – uniquesolution Jun 14 at 19:58
  • $\begingroup$ @uniquesolution Oh, I see what you mean. I forgot to add in to generate the sequence inductively; I just defined the first two terms. Edited, thank you! $\endgroup$ – cmk Jun 14 at 20:18
  • $\begingroup$ Does the sequence need to be constructed inductively? Can one not simply say "for each $n\in\mathbb N$, there exists $x_n\in\mathbb N$ such that $\|Tx_n\|> n^2\|x_n\|$," obtained from the direct negation of the definition of boundedness? Then we get $z_n = \frac{x_n}{n\|x_n\|}$ for all $n$. $\endgroup$ – Luke Collins Jun 14 at 23:48
  • $\begingroup$ I mean, I think that’s fine. I think it’s pretty clear how it’s constructed. I just wrote it out to make sure you understood! $\endgroup$ – cmk Jun 15 at 0:03

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