A question about a dense subset in Banach space.

Question

Let $X,Y$ be Banach spaces, $T:X\to Y$,unbounded linear operator.

How to prove that there is a natural number $n$,the set $\{x:\|Tx\|\le n\|x\|\}$ is dense in $X$?

Answer 1

The claim is true (see the proof below using Baire's Category Theorem). But note that it is (without the axiom of choice) hard to come up with an unbounded operator $T$ that is defined on the whole Banach space $X$.

For $n\in\mathbb{N}$ define $$ A_{n}:=\left\{ x\in X \mid \left\Vert Tx\right\Vert \leq n\cdot\left\Vert x\right\Vert \right\} . $$ Note that $0\in A_{n}$ for all $n\in\mathbb{N}$ and that we have $rA_{n}\subset A_{n}$ for all $r>0$ and $n\in\mathbb{N}$.

Also note that we have $$ X=\bigcup_{n\in\mathbb{N}}A_{n}\subset\bigcup_{n\in\mathbb{N}}\overline{A_{n}}. $$ The Baire Category Theorems shows that there is $n_{0}\in\mathbb{N}$, $x_{0}\in X$ and $\varepsilon>0$ such that $$ B_{\varepsilon}\left(x_{0}\right)\subset\overline{A_{n_{0}}} $$ holds.

Choose $m\in\mathbb{N}$ with $$ \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\varepsilon\right)+\left\Vert Tx_{0}\right\Vert \right]\leq m. $$

Now let $x\in X$ be arbitrary with $\frac{\varepsilon}{2}<\left\Vert x\right\Vert <\varepsilon$, so that $x_{0}+x\in B_{\varepsilon}\left(x_{0}\right)\subset\overline{A_{n_{0}}}$ holds. Thus, there is a sequence $\left(y_{n}\right)_{n\in\mathbb{N}}$ in $A_{n_{0}}$ such that $y_{n}\rightarrow x_{0}+x$ holds. This yields $y_{n}-x_{0}\rightarrow x$ and $$ \begin{eqnarray*} \left\Vert T\left(y_{n}-x_{0}\right)\right\Vert & \leq & \left\Vert Ty_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert \\ & \leq & n_{0}\cdot\left\Vert y_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert \\ & = & \underbrace{\frac{n_{0}\cdot\left\Vert y_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert }{\left\Vert y_{n}-x_{0}\right\Vert }}_{=:\alpha_{n}}\cdot\left\Vert y_{n}-x_{0}\right\Vert .\qquad\left(\ast\right) \end{eqnarray*} $$ Now note that $\frac{\varepsilon}{2}<\left\Vert x\right\Vert <\varepsilon$ yields $$ \begin{eqnarray*} \alpha_{n}\xrightarrow[n\rightarrow\infty]{}\frac{n_{0}\cdot\left\Vert x_{0}+x\right\Vert +\left\Vert Tx_{0}\right\Vert }{\left\Vert x\right\Vert } & < & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left\Vert x_{0}+x\right\Vert +\left\Vert Tx_{0}\right\Vert \right]\\ & \leq & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\left\Vert x\right\Vert \right)+\left\Vert Tx_{0}\right\Vert \right]\\ & < & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\varepsilon\right)+\left\Vert Tx_{0}\right\Vert \right]\\ & \leq & m. \end{eqnarray*} $$ We thus get $\alpha_{n}<m$ for $n$ large enough. The estimate marked with $\left(\ast\right)$ above thus shows $y_{n}-x_{0}\in A_{m}$ for $n$ large enough and thus $x=\lim_{n}\left(y_{n}-x_{0}\right)\in\overline{A_{m}}$.

We have thus established $$ B_{\varepsilon}\left(0\right)\setminus\overline{B_{\varepsilon/2}}\left(0\right)\subset\overline{A_{m}}. $$ This implies $$ B_{r\varepsilon}\left(0\right)\setminus\overline{B_{r\varepsilon/2}}\left(0\right)\subset r\overline{A_{m}}=\overline{rA_{m}}\subset\overline{A_{m}} $$ for all $r>0$. But $$ X\setminus\left\{ 0\right\} =\bigcup_{r>0}\left[ B_{r\varepsilon}\left(0\right)\setminus\overline{B_{r\varepsilon/2}}\left(0\right)\right], $$ so that $X\subset\overline{A_{m}}$ follows. This completes the proof.