it‘s a problem from salamon’s functional analysis
Let $X$ be a real Banach space, Let $Y\subset X$ be a linear subspace, and let $T:Y\to \mathbb{R} $ be a linear functional. show that $T$ is closeable if and only if $T$ is bounded.
The definition of closeable:
Let $X$ and $Y$ be Banach spaces, let $dom(A)\subset X$ be a linear subspace, and let $A:dom(A)\to Y$ be a linear operator. The operator $A$ is called closeable if there exists a closed linear operator $A’: dom(A’)\to Y$ on a linear subspace $dom(A’)\subset X$ such that
$$dom(A)\subset dom(A’), A’x=Ax, for~all~x\in dom(A)$$
If $T$ is bounded, we can easily show that $T$ is closeable by Hahn-Banach theorem and closed graph theorem.
However, conversely, I don’t know how to deal with it. If $T’$ is a closed linear functional which extends $T$, we cannot get the boundedness of $T’$ since the domain of $T’$ may not be closed, and hence we cannot use closed graph theorem. It seems that we should prove that a closed linear functional must be bounded, but I don’t know how to do it.
I do hope someone can help me.Thanks!
Update:
I have some ideas although I’m not sure whether it is right.
Claim: Suppose that $X$ is a normed vector space but not complete, and $T:X\to \mathbb{R}$ is a closed linear functional.We can show that $T$ must be bounded.
Proof: If $T$ is unbounded, then $Ker T$ is a dense subspace of $X$. Chose $y\in X\backslash Ker(T)$, and $\left\{ x_n \right\} \subset Ker\left( T \right) $ such that $x_n\to y$.
Note that $Tx_n\to 0$,$x_n\to y$ implies that $Ty=0 $ since $T$ is closed, so it contradicts with the assumtion that $y\notin Ker(T)$.
Best Answer
You don't need to argue by contradiction, your argument is a direct one.
Indeed, let $\{x_n\}\subset\ker T$ be Cauchy. Since $X$ is Banach, there exists $y\in X$ with $y=\lim x_n$. The sequence $\{(x_n,0)\}$ is a Cauchy sequence in the graph of $T$. As this is closed and $X\times\mathbb R$ is Banach, it is complete; so $(y,0)$ is in the graph of $T$, i.e. $Ty=0$. This shows that $\ker T$ is closed, and so $T$ is bounded (this last implication is where one uses that $T$ is a functional).