TITLE: Unbounded linear operator
QUESTION [7 upvotes]: Let $(A, \|\cdot\|_A), (B, \|\cdot\|_B)$ be normed linear spaces. Consider $T \in L(A,B)$
The operator norm of $T$ is defined to be
$$\|T\| = \sup\{\|Tx\|_B: \|x\|_A \leq 1\}$$
$T$ is bounded if $\|T\| < \infty$ otherwise it is unbounded.
So can someone give me an example of an unbounded linear operator? This seems very counterintuitive to me because, that means 
$$\exists \space x \in A, \|Tx\|_B = \infty$$
but then any scalar multiples of $Tx$ would have an infinite norm. Then what would $T(0)$ be?

REPLY [10 votes]: Take $A=B$ be the set of complex sequences with finitely many nonzero terms:
$$
A=B=\{\,\{x_n\}\,:\ \exists m\in\mathbb N\ \text{ with }x_n=0\ \forall n\geq m\}.
$$
Consider in both the supremum norm ($\|x\|=\max\{|x_1|,|x_2|,\ldots\}$). Define 
$$
T(x_1,x_2,\ldots,x_n,0,0,\ldots)=(x_1,2x_2,3x_3,\ldots,nx_n,0,0,\ldots).
$$
Then $T$ is linear. And, if $e$ is the sequence $(0,\ldots,0,1,0,0,\ldots)$ (the 1 in the $n^{\rm th}$ position), then $\|e\|=1$ and 
$$
\|Te\|=n.
$$
As we can do this for every $n$, $\|T\|=\infty$. 
As you can see, here $\|Tx\|<\infty$ for all $x$. Finally, you ask about $T(0)$; for a linear operator, $T(0)=0$ always (bounded or unbounded, it doesn't matter). 

For a different and maybe more natural example, consider $A=B$ the set of polynomials as a subset of $C[0,1]$, and let 
$$
Tp=p'
$$
be the differentiation operator. This is an unbounded operator, since $\|x^n\|=1$ and $\|T(x^n)\|=n$ for all $n$.