If $V$ and $W$ are topological vector spaces (and $W$ is finite-dimensional) then a linear operator $L\colon V\to W$ is continuous if and only if the kernel of $L$ is a closed subspace of $V$.

Why is this so?

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    $\begingroup$ Since $\ker L = L^{-1}(0)$, one direction is trivial: the singleton $\{0\}$ is closed in $W$, so the continuity of $L$ implies that $\ker L$ is closed in $V$. $\endgroup$
    – Siminore
    Apr 23, 2012 at 13:29
  • $\begingroup$ I see thanks. I have another question If you could perhaps help? Say $M$ is a closed subspace of a Hilbert Space $H$, then we can write $H = M + M^{\perp}$ . Does this hold if M is not closed? $\endgroup$
    – rk101
    Apr 23, 2012 at 13:33
  • $\begingroup$ As far as I know, the orthogonal decomposition is true only for closed subspaces. I mean, true in general. $\endgroup$
    – Siminore
    Apr 23, 2012 at 13:34
  • $\begingroup$ @rk101 For your second question, maybe you should open an other one. And the answer is no if the Hilbert space is infinite dimensional, since you can find a non-continuous linear functional. Its kernel is a dense strict subspace, so its orthogonal is $\{0\}$. $\endgroup$ Apr 23, 2012 at 13:44
  • $\begingroup$ I don't think your solution is corrected. Why do you make sure that" {wn} can't have any convergent subsequence, and kerL is not closed". In this case, KerL is not compact, so it doesn't require that every sequence in a closed subset must be have convergent subsequence. $\endgroup$ Mar 27, 2013 at 1:27

2 Answers 2


You need the hypothesis that $V$ and $W$ be Hausdorff, although this is usually given for granted.

One direction is as in Siminore's comment.

For the other, since $\text{ker}L$ is closed, the quotient space $V/\text{ker}L$ is Hausdorff. Algebraically $V/\text{ker}L\cong\text{im}L,$ and $\text{im}L$ is a subspace of $W.$ The algebraic isomorphism $V/\text{ker}L\cong\text{im}L$ derived from $L$ is a topological isomorphism because finite-dimensional spaces admit only one Hausdorff topological vector space topology (any linear isomorphism between finite-dimensional Hausdorff topological vector spaces is a topological isomorphism). This means that $L$ is continuous, as a composition of the quotient map $V\to V/\text{ker}L$ with the isomorphism between $V/\text{ker}L$ and $\text{im}L$ with the inclusion $\text{im}L\to W.$

  • 1
    $\begingroup$ You only need $W$ to be Hausdorff for this argument to work. $\endgroup$
    – t.b.
    Apr 23, 2012 at 14:12

Assume $W=\mathbb{R}$ for simplicity. If $L$ is not continuous at zero, there is a sequence $\{v_n\}$ in $V$ such that $|v_n|=1$ and $|Lv_n| \to +\infty$. Assume that $L \neq 0$, the null map. Then you can fix $z \in V$ such that $Lz=1$. Consider now $w_n=v_n-(Lv_n)z \in V$. Trivially, $Lw_n=0$, so that $w_n \in \ker L$. But $$|w_n| \geq \left| |L v_n| |z| - |v_n| \right| \to +\infty.$$ Hence $\{w_n\}$ can't have any convergent subsequence, and $\ker L$ is not closed. This is the proof if $V$ has a norm. In the general case, the reasoning is rather similar, but one needs to know the concept of balanced neighborhood. A precise reference is Theorem 1.18 of Rudin, Functional Analysis.

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    $\begingroup$ What does $|v_n|$ mean if $v_n\in V$, a topological vector space which may be not a normed space? $\endgroup$ Apr 23, 2012 at 13:45
  • $\begingroup$ You are right, and I'm not very familiar to topological vector spaces without a (semi)norm. The proof follows similar lines, but the unit ball is replaced by a fixed neighborhood of zero. $\endgroup$
    – Siminore
    Apr 23, 2012 at 14:01
  • 2
    $\begingroup$ Wouldn't your argument about a convergent subsequence just imply that ker L is not compact, not open? $\endgroup$
    – asmeurer
    Feb 12, 2013 at 3:41
  • 2
    $\begingroup$ How is showing an unbounded sequence exists in a set proving this set being closed? There are closed sets that contain unbounded sequences. Please explain. $\endgroup$
    – xzhu
    Dec 11, 2013 at 18:32
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    $\begingroup$ Just take your $w_n = z - \frac{v_n}{L(v_n)}$, then $f(w_n) = 0$ for all $n$, but $w_n \rightarrow z$ which is not in the $\ker L$. Therefore $\ker L$ is not closed. $\endgroup$
    – Xiao
    Sep 4, 2014 at 15:08

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