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Let $X,Y$ be normed spaces and $T:X\to Y$ linear and discontinuous.

Hence $T$ is discontinuous at every point. Then for every $x\in X$ there exists a sequence $(x_n)\subseteq X$ such that $(x_n)$ converges to $x$ and $(T(x_n))$ doesn't converge to $T(x)$.

My question is:

Can we choose above a sequence $(x_n)$ such that $(T(x_n))$ is bounded?

Since $(x_n)$ converges to $x$ then $(x_n)$ is bounded. But since $T$ is discontinuous it sends every ball to an unbounded set, so I'd like to know if such a sequence can be found carefully.

Any ideas? Thank you.

Tanius
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1 Answers1

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Since $T$ is discontinuous at $0$, we can pick a sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\rightarrow 0$ and $Tx_n\not \rightarrow 0$. By passing to a subsequence we may also assume that $Tx_n\neq 0$ for every $n$. If $(Tx_n)_{n\in\mathbb{N}}$ is unbounded then set $y_n=\frac{x_n}{\|Tx_n\|}$. Clearly, $y_n\rightarrow 0$ and $\|Ty_n\|=1$ for every $n$, so the sequence $(Ty_n)_{n\in\mathbb{N}}$ is bounded and it cannot converge to $0$ as all of its terms have norm one.

For the general $x\in X$, consider the sequence $z_n=y_n+x$.

tree detective
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