2

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!

Community
  • 1
Tim
  • 47,382

2 Answers2

7

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.

azarel
  • 13,120
2

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.

Martin Argerami
  • 205,756
  • This might be too late. But I don't think this proof is correct. Because for a general cts linear operator $f: X\to Y$, it's not guaranteed $f$ is bounded on some neighborhood of the origin. To show a cts $f$ maps bdd sets to bdd sets, it's enough to use the fact that preimage of neighborhoods in Y are neighborhoods in X, and that neighborhoods are absorbing. – user760 Apr 04 '23 at 19:28
  • Proof. Suppose $f: X\to Y$ is cts. Let $A\subset X$ be bdd. Then, for every neighborhood $W\ni 0_Y$ in $Y$, $f^{-1}(W)\ni 0_X$ is a neighborhood in $X$. Then there exists $t>0$ s.t. $t A\subset f^{-1}(W)$, and so $t f(A)=f(tA)\subset f\left(f^{-1}(W)\right)=W$. Since $W$ is arbitrary, $f(A)$ is then a bdd set in $Y$. $\square$ – user760 Apr 04 '23 at 21:18
  • 1
    Fair enough. I think (who knows, it was 11 years ago) that I was trying to answer 'does "$f(V)$ is bounded for some neighborhood $V$ of $0$" imply that $f$ maps bounded subsets to bounded subsets?" ' – Martin Argerami Apr 04 '23 at 21:51