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?
Asked
Active
Viewed 1.8k times
22
Davide Giraudo
- 172,925
rk101
- 337
-
6Since $\ker L = L^{-1}(0)$, one direction is trivial: the singleton ${0}$ is closed in $W$, so the continuity of $L$ implies that $\ker L$ is closed in $V$. – Siminore Apr 23 '12 at 13:29
-
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? – rk101 Apr 23 '12 at 13:33
-
As far as I know, the orthogonal decomposition is true only for closed subspaces. I mean, true in general. – Siminore Apr 23 '12 at 13:34
-
@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 ${0}$. – Davide Giraudo Apr 23 '12 at 13:44
-
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. – Hoang Nguyen Mar 27 '13 at 01:27
2 Answers
11
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 $\text{im}L\to W.$
Xabier Domínguez
- 1,033
5
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.
Siminore
- 35,136
-
3What does $|v_n|$ mean if $v_n\in V$, a topological vector space which may be not a normed space? – Davide Giraudo Apr 23 '12 at 13:45
-
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. – Siminore Apr 23 '12 at 14:01
-
You don't need to assume that $L \neq 0$. If $L = 0$, then it is automatically continuous, and the kernel, the whole space, is closed. – asmeurer Feb 12 '13 at 03:19
-
2Wouldn't your argument about a convergent subsequence just imply that ker L is not compact, not open? – asmeurer Feb 12 '13 at 03:41
-
2How is showing an unbounded sequence exists in a set proving this set being closed? There are closed sets that contain unbounded sequences. Please explain. – xzhu Dec 11 '13 at 18:32
-
-
14Just 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. – Xiao Sep 04 '14 at 15:08