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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?

glS
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Phantom
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  • Have you seem some of the examples in the related links? http://math.stackexchange.com/questions/344623/show-that-linear-operator-on-ell2-is-unbounded?rq=1 for example. – Sarvesh Ravichandran Iyer Sep 27 '16 at 23:29
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    Derivation operator $D$ on $(C_c^{\infty}(\Bbb{R}), |\cdot|_{\sup})$ given by $Df = f'$ is a qunitessential example of unbounded linear operator. – Sangchul Lee Sep 27 '16 at 23:45

1 Answers1

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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 (though it is intrinsically the same), 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 $n$.

Martin Argerami
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