# Sequential Continuity of Linear Operator implies Boundedness

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, $$\|Tx_N\| = \|T\tfrac{x}{N}\| = \frac1N\|Tx\| < \|x\| \implies \|Tx\| so $$T$$ is bounded.

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

## 1 Answer

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 $$Tz_n\not\rightarrow T(0)=0.$$

• How exactly is the sequence $x_n$ defined? – uniquesolution Jun 14 at 19:58
• @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! – cmk Jun 14 at 20:18
• 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 $n$. – Luke Collins Jun 14 at 23:48
• 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! – cmk Jun 15 at 0:03