# Is there an example of a linear map between Banach Spaces that is not closed?

Is there a simple example of a linear map $$T: X \to Y$$ between Banach spaces that is not closed, i.e. if $$x_n \to 0$$ in $$X$$ and $$T(x_n) \to y$$ in $$Y$$, then $$y=0$$? I know the Closed Graph Theorem says that if $$T: X \to Y$$ is a linear map between Banach spaces, then \begin{align*} T \text{ is bounded } \iff T \text{ is closed } \iff \text{gr}(T) \text { is a closed subspace of } X \oplus Y. \end{align*} So I would have to find an unbounded linear map between Banach spaces, correct? What exactly would that be?

We can construct an unbounded linear functional easily. Pick your favorite basis $$e_n$$ with $$\lVert e_n \rVert = 1$$ of $$X$$ (which can be done with the axiom of choice) and define $$f$$ as $$f(x)=\sum_n n x_n,\,\,\,\,\text{where }\,\,\,\,x=\sum_n x_n e_n.$$ Note that this is well-defined, as $$x_n$$ is non-zero for finitely many $$n$$ for every $$x$$. Then pick $$y_n = \frac{e_n}{\sqrt{n}},$$ which clearly goes to $$0$$, but $$f(y_n)=\sqrt{n} \to +\infty$$.
• What exactly would $X$ and $Y$ be? Are there any concrete Banach spaces that this works for? Mar 1, 2021 at 21:37
• @MATH-LORD Pick your favorite Banach space $X$ over the field $\mathbb{F}$ and $Y=\mathbb{F}$. Note that you can't construct this function "exactly", but you can prove that it exists. Mar 1, 2021 at 21:39