3

If $E$ is a topological vector space (TVS), $F_1$ a closed subspace of $E$, and $F_2$ a finite dimensional subspace of $E$, such that $F_1 \cap F_2=\{0\}$, is $F_1+F_2$ necessarily closed? If yes, are the projection from $F_1+F_2$ onto $F_1$ and $F_2$ respectively, continuous?

Jim
  • 31

2 Answers2

3

Yes If we suppose that $E$ is a normed space , the sum of a closed subspace $Y$ and a finite dimensional space $F$ is closed, in fact : suppose that a sequence $(y_n+f_n) \subset Y+F$ converge vers $x\in X$, Let $P$ the (continuous Why? ) projection from $Y+F$ to $F$, the sequence $(y_n+f_n)$ is bounded (because it converge to $x$ so it exist a subsequence $(f_{n_k})$ that converge to $f\in F$ (because F is finite dimensional vector space so every bounded closed set is compact) and so $y_{n_k}$ converge to $y\in Y$ as difference of two convergent sequences, and so $x=y+f\in Y+F$

To complete the proof we need to proof that the projection from $Y+F$ to $F$ is continuous, so for that we put $$ \delta=\min\{d(f,Y) : \|f\|=1 \} $$ $\delta>0$ because the set $S_F=\{ f\in F ; \|f\|=1\}$ is compact, and the function $d(.,Y)$ is a continuous function so it exist $f'\in S_F$ such that $\delta=d(f',Y)$, then if $\delta=0$ this implies that $f'\in Y\cap F=\{0\}$ but $\|f'\|=1\neq 0$, absurde.

so $$ \|y+f\|\geq \delta\|f\| $$ so the projection from $Y+F$ to $F$ is of norme $\leq \delta^{-1}$

and the projection from $Y+F$ to $Y$ is of norme $\leq 1+\delta^{-1}$

Hamza
  • 3,831
  • 3
    The sum of two closed subspaces is not necessarily closed. http://math.stackexchange.com/questions/135471/the-direct-sum-of-two-closed-subspace-is-closed-hilbert-space Clealrly that one of the subspaces is finite dimensional must be important. – Markus Aug 13 '15 at 01:05
  • yeah that true I will correct the proof – Hamza Aug 13 '15 at 01:19
  • The proof assumes continuity of the projection. As far as I remember, however the subtlty of closed subspaces is equivalent to the spaces being closed. That might be the flaw in the proof. – freishahiri Aug 13 '15 at 20:46
  • no I proof the continuity of projection on the second part, for this I need the normed structure. @Freeze_S – Hamza Aug 13 '15 at 23:08
  • I think you have to explain better why $\delta>0$, that's the crux of the entire argument. If $\delta=0$, the argument fails. – Markus Aug 14 '15 at 22:08
  • @Markus I think it's clear now :) – Hamza Aug 15 '15 at 17:41
  • @Hamza Yes, and compactness of the unit ball is why it works when one space is finite dimensional. That inf can be zero when both spaces are infinite dimensional. – Markus Aug 16 '15 at 13:08
1

Whenever N,F are subspaces of a TVS, N closed and F finite-dimensional, N+F is closed. This is Theorem 1.42 (p. 30) in Rudin: Functional Analysis, 2nd edition, 1991. https://archive.org/details/RudinW.FunctionalAnalysis2e1991

But even if the TVS is Hilbert, F being closed is not enough (By Exercise 1.20, p. 40), it must be finite-dimensional.

Theorem 5.16b (p. 126): If N and F are closed subspaces of an F-space N+F, and $N\cap F=\{0\}$, then the projection $P:A+B\to A$ with null space $B$ is continuous.

So to your first question the answer is "yes", to the second "yes if N+F is an F-space" but not in general, by Exercise 5.9.

Exercise 5.9 (p. 145): Here $N,F$ are closed subspaces over $L^2(0,1)$, $N\cap F=\{0\}$ but yet $P$ is not continuous.

Note: it is not enough that N,F are closed subspaces of the F-space $L^2$; the space $N+F$ should be complete (F-space).

Def. 1.8e: F-space = TVS whose topology is induced by a complete invariant metric (equivalently, TVS having a complete N1 topology, by Theorem 1.24)

BTW, only in Theorem 1.42 finite-dimensionality is assumed, not in the others. In Rudin, TVSs are Hausdorff.

user3810316
  • 735