Question: Let $E$ be a normed space. Let $G$ be a closed subspace of $E$ and let $F$ be a finite dimensional subspace of $E$. Show that $F+G$ is a subspace of $E$ and is closed.
I'm having trouble in showing $F+G$ to be closed. I know that $F$ is itself closed and complete, as it is a finite dimensional subspace of a normed space, and that if $F$ were compact that $G+F$ would be closed. I also know that the closed unit ball of any finite dimensional normed space is compact.
I tried two methods. One was to take a convergent sequence $(x_n)_{n \in \mathbb N}$ in $G+F$. Then we can write $x_n = f_n + g_n$ where $f_n$ and $g_n$ are sequences in $F$ and $G$ respectively, and I attempted to find a way to force the individual components $f_n$ and $g_n$ inside the unit ball which would enable me to say that they had convergent subsequences. I couldn't see how to do this, however.
The other thought was to try and show that $F$ is compact, but I don't see a way to do this as I can't imagine it to be simply true without some other conditions on $F$.
Are one of these methods the right way to go? Or should I go another direction? I would appreciate any help I can get, although I would prefer not to be presented with a full proof so that I can do some work for myself. Thanks!
Best Answer
The easiest way to prove it is to consider the quotient $E/G$. Let $\pi$ be the canonical map. Then $\pi(F) \subset E/G$ is finite-dimensional, hence closed ($E/G$ is Hausdorff since $G$ is closed), and $F + G = \pi^{-1}(\pi(F))$ is closed as the preimage of a closed set in $E/G$.
An alternative method:
We can without loss of generality assume that $F \cap G = \{0\}$. Otherwise let $F'$ be a complement of $F\cap G$ in $F$ and consider $F'$ instead of $F$. Since $F' + G = F + G$ the result is unaffected.
Consider the space $H = F \oplus G$ with the norm $\lVert (f,g)\rVert_H = \lVert f\rVert_E + \lVert g\rVert_E$. It is easy to see that $\alpha \colon H \to E;\; \alpha((f,g)) = f+g$ is continuous. Using the finite-dimensionality of $F$, it is not hard to show that $\alpha$ is an embedding, there is a $\delta > 0$ with
$$\lVert f + g\rVert_E \geqslant \delta\cdot \lVert (f,g)\rVert_H.$$
(If there weren't, you'd have a sequence $p_n$ with $\lVert p_n\rVert_H = 1$ and $\lVert \alpha(p_n)\rVert_E \to 0$. Use the finite-dimensionality of $F$ to obtain a contradiction.)
Then, if $x \in \overline{F+G}$, you have a sequence $x_n \in F+G$ converging to $x$. Using the fact that $\alpha$ is an embedding, conclude that $x \in F+G$.