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?

  • $\begingroup$ 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). $\endgroup$ – Jochen Mar 22 at 12:53
  • $\begingroup$ That's indeed what I meant, thanks for pointing that out. $\endgroup$ – 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 $X$.

  • $\begingroup$ thanks jmerry that's great. $\endgroup$ – Dan Mar 22 at 17:44

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