# “A linear operator $f$ is continuous if $f(V)$ is bounded for some neighborhood $V$ of $0$.”

From Wikipedia

a linear operator $f$ between two topological vector spaces is continuous if $f(V)$ is bounded for some neighborhood $V$ of $0$.

1. I wonder why it is true?

If I understand correctly, "$f(V)$ is bounded for some neighborhood $V$ of $0$" is same as saying $f$ is locally bounded?

2. I saw elsewhere that for a linear operator between two TVSes, continuity implies mapping bounded subsets to bounded subsets.

So if $f$ is linear mapping between two TVSes, does "$f(V)$ is bounded for some neighborhood $V$ of $0$" imply that $f$ maps bounded subsets to bounded subsets?

Thanks and regards!

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Why don't you read Rudin's functional analysis? Everything you asking about TVS last time is explained there. –  Norbert Feb 27 '12 at 6:52
@Norbert: okay! –  Tim Feb 27 '12 at 6:55

Let $f: X\to Y$ be a linear mapping between two TVS with the property that $f(V)$ is bounded for some neighbourhood $V$ of $0$. It suffices to show that $f$ is continuous at $0$, let $U$ be a neighbourhood of $0$ at $Y$. By definition of being bounded there is a $t$ so that $f(V)\subseteq t\cdot U$ this implies $\frac 1 t\cdot f(V)\subseteq U$. Since $\frac 1 t\cdot f(V)=f(\frac 1 t \cdot V)$ and $\frac 1 t \cdot V$ is open we obtain that $f$ is continuous.
To answer your second question: yes, a continuous function will map bounded sets to bounded sets. Assume $f:X\to Y$ has a neighbourhood $V$ of $0$ with $f(V)$ bounded. If $A\subset X$ is bounded, there exists a number $t$ with $tA\subset V$. Then $tf(A)=f(tA)\subset f(V)$. Now fix any neighbourhood of $0$ in $Y$, say $W$. Since $f(V)$ is bounded, there exists $s$ with $s f(V)\subset W$. Then $$st f(A) = s f(tA) \subset s f(V)\subset W.$$ We conclude that $f(A)$ is bounded.