# $X,Y$ Banach, $V \subset X$ a linear subspace, $T:V\to Y$ closed, then $T$ bounded $\iff$ $V$ closed

Let $$X$$ and $$Y$$ be Banach spaces and let $$V \subset X$$ be a linear subspace. Let $$T: V \to Y$$ be a closed linear operator. Show that $$T$$ is bounded if and only if $$V$$ is closed.

For the direction $$V$$ closed $$\Rightarrow$$ $$T$$ bounded this is simply the closed graph theorem. However I am stuck for the direction $$T$$ bounded $$\Rightarrow$$ $$V$$ closed.

I tried to prove $$V$$ not closed $$\Rightarrow$$ $$T$$ unbounded, by something along the lines of

let $$v_n$$ be a sequence in $$V$$ converging to a limit $$x \in X\setminus V$$ then $$(v_n,Tv_n)$$ has limit $$(x,y)$$ for $$y=\lim Tv_n$$ and then I guess we seek to show something like that $$\frac{||y||}{||x||}$$ is unbounded if we choose $$v_n$$ in the right way. Can anyone help?

• Probably, you mean be the closedness of $T$ that its graph $\{(x,T(x)):x\in V\}$ is not only closed in $V\times Y$ but even in $X\times Y$ (this is not the standard terminology). – Jochen Mar 22 at 12:53
• That's indeed what I meant, thanks for pointing that out. – Dan Mar 22 at 17:41

The way I'd do it: Assume $$T$$ is a bounded linear operator. Let $$x$$ be a point in the closure of $$V$$, and let $$v_n$$ be a Cauchy sequence of points in $$V$$ converging to $$x$$. Then, since $$T$$ is uniformly continuous ($$\|Tx-Ty\|\le \|T\|\cdot\|x-y\|$$), $$Tv_n$$ is a Cauchy sequence in $$Y$$. Let its limit be $$y$$.
Then the points $$(v_n,Tv_n)$$ converge to $$(x,y)$$ in $$X\times Y$$. Since the graph of $$T$$ is closed in $$X\times Y$$ by hypothesis, $$(x,y)$$ is in the graph. In particular, $$x\in V$$. This works for any $$x$$ in the closure, so $$\overline{V}\subset V$$ and $$V$$ is closed in $$X$$.