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Let $X,Y$ be normed linear spaces. Let $T: X\to Y$ be linear. If $X$ is finite dimensional, show that $T$ is continuous. If $Y$ is finite dimensional, show that $T$ is continuous if and only if $\ker T$ is closed.

I am able to show that $X$, finite dimensional $\implies$ $T$ is bounded, hence continuous.

For the second part: This is what I have:

Suppose $T$ is continuous. By definition $\ker T = \{ x\in X : Tx = 0 \}$ , and so $\ker T$ is the continuous inverse of a closed set. Hence $\ker T $ is closed.

First, is what I have attempted okay. How about the other direction?

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  • $\begingroup$ What kind of linear spaces are we talking about? I'm assuming they're over $\mathbb{R}$ or $\mathbb{C}$. If $X$ is infinite-dimensional then it is not clear what topology to use, unless $X$ is e.g. an inner product space. $\endgroup$ Apr 14 '12 at 21:00
  • $\begingroup$ What you have so far looks good. $\endgroup$ Apr 14 '12 at 21:08
  • $\begingroup$ Is it possible to say something even if we assume $X$ to be infinite-dimensional (e.g. a Banach or Hilbert space)? $\endgroup$
    – ric.san
    Jun 10 at 15:03
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If $\ker(T)$ is closed then $X/\ker(T)$ is a normed vector space. Observe that the map $\overline{T}:X/\ker(T)\to Y$ given by $\overline{T}(x+\ker(T))=T(x)$ is a well-defined linear map by the first part $\overline T$ is continuous since $X/\ker(T)$ is a finite dimensional vector space (as it is isomorphic to a subspace of $Y$). Let $\pi: X \to X/\ker(T)$ denote the quotient map. Note that $T=\overline{T}\circ \pi$ hence $T$ is continuous since it is a composition of continuous functions.

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    $\begingroup$ It remains to prove that all linear mappings between finite-dimensional normed spaces are continuous. IMHO, this is something that looks more trivial than it really is: see math.stackexchange.com/a/64756/8157 and comments therein. $\endgroup$ Apr 14 '12 at 21:29
  • $\begingroup$ @GiuseppeNegro The OP already proved that. It was the first part of his problem list. $\endgroup$
    – azarel
    Apr 14 '12 at 21:35
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    $\begingroup$ $X/ker T$ is isomorphic to Im T, not necessarily Y. $\endgroup$
    – plm
    Apr 15 '12 at 15:34
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    $\begingroup$ @plm True, it is still finite dimensional though. $\endgroup$
    – azarel
    Apr 15 '12 at 18:33
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    $\begingroup$ Awesome proof!! $\endgroup$
    – Error 404
    Jan 12 '18 at 8:11
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I am doing for $Y=\mathbb{R}$; Clearly if $f$ is continuous then its kernel is closed set. for the converse, assume that $f\neq0$ and that $f^{-1}(\{0\})$ is a closed set. Pick some $e$ in $X$ with $f(e)=1$. Suppose by way of contradiction $||f||=\infty$. Then there exists a sequence $\{x_n\}$ in $X$ with $||x_n||=1$ and $f(x_n)\ge n$ for all $n$. Note that the sequence $\{y_n\}$ defined by $y_n=e-\frac{x_n}{f(x_n)}$, satisfies $y_n\in f^{-1}(\{0\})$ for all $n$ and $y_n\rightarrow e$. Since the set $f^{-1}(\{0\})$ is closed it follows that $e$ must belong to it and consequently $f(e)=0$ which is a contradiction. Thus $f$ is a continuous linear functional.

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    $\begingroup$ Can you explain how you generalise to $Y$ finite dimensional ? $\endgroup$
    – user10676
    Apr 15 '12 at 12:05
  • $\begingroup$ If you say that $y_n\to 0$ as $n\to\infty$ you are already assuming that $x_n\to x$ and $f(x)=0$, which implies that $x\in\ker f$ and $\ker f$ is closed. Am I right? $\endgroup$ Apr 29 '18 at 18:06

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