Showing the existence of at least one unbounded linear operator

Question

I'm trying to prove the following:

Let $(X, ||\cdot||_X)$ and $(Y, ||\cdot||_Y)$ be normed spaces with $\dim X = \infty$ and $Y \ne \{0\}$. Show that there exists at least one unbounded linear operator $T : X \rightarrow Y $.

I think I have to use the fact that there is a Hamel basis but am not sure how to start. Can anyone start me off?

Answer 1

Let $\{b_i: i \in I\}$ be an infinite Hamel basis for $X$ ( I will assume WLOG that $\mathbb{N} \subseteq I$) and pick $y \neq 0$( say with $\|y\|=1$ WLOG) in $Y$. Define $f(b_i) = i\cdot y$ when $i \in \mathbb{N}$ and $f(b_i) = 0$ otherwise, and extend by linearity.

$f$ is then unbounded, as is easy to see.

Answer 2

Since $X$ is infinite dimensional we have countable set $\{x_n\}_{n\in \Bbb N}$ which is linearly independent. Define a map $T\colon X\rightarrow Y$ as follows $T\left(\frac{x_n}{||x_n||}\right)=ny$ and then extend $T$ linearly, where $y\in Y$ is a fixed nonzero element and $T(x)=0$ if $x\not \in \operatorname{span}\{x_n\}$, i.e., $T$ is zero on the complement of $\operatorname{span}\{x_n\}$. This map is well defined and linear since $\left\{\frac{x_n}{||x_n||}\right\}$ is a linearly independent subset of $X$.

Now, $T$ is not bounded since $\left\{\frac{x_n}{||x_n||}\right\}_n\subseteq \{x\in X: ||x||≤1\}$ and $$\sup_{x\in \{x\in X: ||x||≤1\}} \ ||T(x)||≥n||y||>0$$ for each $n\in \Bbb N$.