Let $X$ be a Banach space. Suppose $A:D(A)\subset X \longrightarrow X$ is a closed linear operator and $D(A)$ is dense in $X$. Prove that $A$ cannot be extended as a closed linear operator to any subspace $\mathcal{D} \supset D(A)$. The density of $D(A)$ is needed, since the restriction of a closed linear operator to a closed subspace is still a closed linear operator. Any hint is appreciated. Thanks.
Functional Analysis – Extension of Densely Defined Closed Operators
banach-spacesdense-subspacesfunctional-analysistopological-vector-spaces
Related Solutions
Let $P$ be the vector space of polynomials; since each polynomial is a finite linear combination of monomials and $T(x^k) = 0$ for all $k$, $T : P \to P$ is the zero operator. In particular, $T$ is continuous with respect to the sup norm over $[0,1]$ (constant functions are always continuous).
Now recall that $P$ is dense in $C([0,1])$. Clearly the zero operator is a continuous extension of $T$. Suppose $T'$ is another continuous extension of $T$. It's a standard fact from topology that if two continuous functions taking values in a Hausdorff space agree on a dense set, they are the same function. (In this case, note that $T-T'$ is continuous, therefore $(T-T')^{-1}(\{0\})$ is closed...) So this $T$ has a unique continuous extension to all of $C([0,1])$: namely, the zero operator.
More generally, if $X$ is a normed space, $Y$ is a Banach space, $E$ is a dense subspace of $X$, and $T : E \to Y$ is continuous, then there exists a continuous extension $T' : X \to Y$. This is often called the BLT theorem (for "bounded linear transformation"). By the argument we gave above, once the extension exists, it is necessarily unique.
Hahn-Banach covers the complementary case where $E$ is not dense. In this case, in general you are only guaranteed a continuous extension if $Y$ is finite dimensional, and the extension is never unique.
Let $A: D(A): \to X_1$ be some closed operator between Banach spaces and let $x\in X_0$ be some element not in $D(A)$. Define $$A': D(A)+\Bbb C \cdot x\to X_1\oplus_{\ell^1} \Bbb C, \qquad y+ \lambda x\mapsto (A(y), \lambda).$$
We will check that this is also a closed operator. Suppose $y_n +\lambda_n x$ converges and $(A(y_n), \lambda_n)$ also converges. It follows then that $\lambda_n$ converges (call the limit $\lambda$) and also that $A(y_n)$ converges. From $\lambda_n$ and $y_n+\lambda_n x$ converging you get that $y_n$ converges, call the limit $y$. By closedness of $A$ you get that $y\in D(A)$ and $A(y_n)\to A(y)$. So $$A'(y_n+\lambda_n x) = (A(y_n), \lambda_n) \to (A(y),\lambda) =A'(y+\lambda x)$$ verifying that $A'$ is closed.
If $i: X_1\to X_1\oplus_{\ell^1}\Bbb C$ is the inclusion note that $i\circ A$ is a closed operator. Putting it all together you have that $A'$ is a proper closed extension of the closed operator $i\circ A$. In particular $D(i\circ A) = D(A)\subseteq D(A')$ is densely defined, but $\overline{i\circ A}= i\circ A\neq A'$. As such $D(A)$ is not a core for $A'$.
Best Answer
A counterexample to your claim on Hilbert spaces would be the closure $\bar{A}$ of any non-self-adjoint but symmetric, densely defined and not bounded operator $A$ admitting a self-adjoint extension (which exists because you can construct symmetric operators with any two deficiency indices if I recall correctly, so in particular you can take $A$ with deficiency indices $(1,1)$). The not bounded part is optional, it's just so that we know the self-adjoint extension can't be bounded either, and thus the domain cannot be the whole space, in order to have a non-trivial example. Said self-adjoint extension is closed and thus must extend the aforementioned closure $\bar{A}$ at least, which exists because all symmetric operators are closable.
For a clear-cut example, the following operator from Example 1 in Chapters X of Reed and Simon's Fourier Analysis, Self-Adjointness (volume 2 of their series of books) has deficiency indices $(1,1)$: $$T : f \in H^1_0(0,1) \subset L^2(0,1) \mapsto i f' \in L^2(0,1)$$ whose adjoint is still $f \mapsto if'$ but with domain $H^1(0,1)$.