Showing that the graph of a continuous linear mapping is closed if the domain is closed

Question

I'm working my way through the book Introduction to Hilbert Spaces with Applications and trying to do the exercise of proving a theorem (1.5.11) (its proof is left as an exercise).

It states that if $L: E_1 \rightarrow E_2$ is a continuous linear mapping, then:

• The nullspace $\mathcal{N}(L)$ is a closed subspace of $E_1$
• If the domain $\mathcal{D}(L)$ is closed then the graph $\mathcal{G}(L)$ is a closed subspace of $E_1 \times E_2$

The first statement was pretty straight forward, but I'm having a few concerns about my solution (bottom) to the second and would like to ask for help with:

1. How does the choice of norm on $\mathcal{G}(L)$ affects the answer? The book has yet to suggest one for that space and I used an alternative I found elsewhere. But is this statement true regardless of the norm chosen and if so how can I show that?
2. Does my solution look reasonable?

Thanks!

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Since $E_1$ and $E_2$ were not specified at all I assumed they're both normed spaces for continuity to be definable. Furthermore, given $(x, y) \in E_1 \times E_2$, I take the norm of $E_1 \times E_2$ to be $||(x, y)||_{1, 2} = ||x||_1 + ||y||_2$.

To prove the statement I took the path of showing that $\mathcal{G}(L)$ is closed by asserting that every sequence $(z_n)$ in $\mathcal{G}(L)$ with $z_n \rightarrow z$, implies $z \in \mathcal{G}(L)$. Take any such sequence. By the definition of the graph, for every $n$ there exists $x_n \in \mathcal{D}(L)$ such that

\begin{equation} z_n = (x_n, Lx_n) \end{equation}

Taking $z = (x, \zeta)$ we have $(x_n, Lx_n) \rightarrow (x, \zeta)$, or \begin{equation} \forall \varepsilon > 0, \, \exists N \in \mathbb{N} \text{ such that } ||(x_n, Lx_n) - (x, \zeta)||_{1, 2} = ||x_n - x||_1 + ||Lx_n - \zeta||_2 < \varepsilon \ \forall \, n \ge N \,. \end{equation}

This implies $x_n \rightarrow x$ and $Lx_n \rightarrow \zeta$. Since $\mathcal{D}(L)$ closed, $x \in \mathcal{D}(L)$. Also, because $L$ is continuous, $\zeta = Lx$. Therefore, $z \in \mathcal{G}(L)$. $$\tag*{$\Box$}$$

Answer 1

Your solution looks good. In regards to the choice of norm on the graph I believe that unless the operator is bounded then the only appropriate (in the sense of convergence) choice of norm is the graph norm itself. Notice that, \begin{align} \|x\|_{E_{1}}\leq\|x\|_{E_{1}}+\|Tx\|_{E_{2}}=:\|x\|_{T}, \end{align} and if $T$ is not bounded then the graph norm is stronger than the norm on $E_{1}$. Now clearly you have that $T$ is bounded, but if $T$ is not bounded then convergence in $\|\cdot\|_{E_{1}}$ implies nothing about convergence in $\|\cdot\|_{T}$.

As you may have worked out however, is if $T$ is bounded then $\|\cdot\|_{E_{1}}$ and $\|\cdot\|_{T}$ are equivalent norms. In which case if, \begin{align} \|x_{n}-x\|_{E_{1}}\to0, \end{align} then, \begin{align} \|Tx_{n}-\zeta\|_{E_{2}}\to0. \end{align}

The point here is that what you have shown ($T$ continuous $\implies$ graph$(T)$ closed) is in fact an if and only if statement (please do not take what I have written to be a proof, it is not).