Show that $F+G$ is closed when $G$ a closed subspace of normed space $E$ and $F$ a finite dimensional subspace of $E$.

Question

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!

Answer 1

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 $x \in F+G$.

Answer 2

Another approach:

Start with the case where $F$ is one-dimensional: $F = \mathbb{R}x$. If $x \in G$ there is nothing to show, so suppose $x \notin G$. By the Hahn-Banach theorem, there is a continuous linear functional $f$ with $f(G) = 0$ and (after rescaling) $f(x)=1$. Now suppose $y_n := g_n + c_n x \to y$. We have $c_n = f(y_n) \to f(y)$ so $c_n$ converges to $c = f(y)$. Then $g_n = y_n - c_n x \to y - cx$. $G$ is closed so $y - cx \in G$, hence $y = (y-cx) + cx \in G + \mathbb{R}x$.

For the general case, proceed by induction on the dimension of $F$.

I feel somehow like Hahn-Banach may be buried within Daniel Fischer's answer, but I don't see it so I may be wrong.

Answer 3

I have a more constructive proof:

We use the following fact:

Let $Y$ be a subspace of a normed vector space $X$. Then $Y$ is closed if and only if $Y$ with the restricted metric is complete.

Our proof is the following:

Obviously, $G+F$ is a subspace of $E$. To proof that it is closed, we proceed by induction. Therefore we can asumme without restriction that $\dim N=1$. Let $z\in E$ such that $F:=\left\{\lambda z : \lambda \in \mathbb{K} \right\}$ ($\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$) and $(x_{n})_{n\in\mathbb{N}}=(g_{n}+\lambda_{n}z)_{n\in\mathbb{N}}$ a Cauchy sequence in $G+F$. Then we have two cases:

1. $(\lambda_{n})_{k\in\mathbb{N}}$ is bounded. Then it contains a convergent subsequence $(\lambda_{n_{k}})_{k\in\mathbb{N}}$. Then the sequence $(g_{n_{k}})_{k\in\mathbb{N}}=(x_{n_{k}}-\lambda_{n_{k}})_{k\in\mathbb{N}}$ is convergent. If it is not obvious, the proof is as follows:

   Because $(\lambda_{n_{k}})_{k\in\mathbb{N}}$ is convergent, $(\lambda_{n_{k}})_{k\in\mathbb{N}}$ is Cuachy. Let $\varepsilon>0$, then there is $N_{1}$ such that if $k,s\geq N_{1}$, then $$\left|\lambda_{n_{k}}-\lambda_{n_{s}}\right|<\frac{\varepsilon}{2\left\|z\right\|}.$$ For other hand, as $(x_{n})$ is cauchy, there is $N_{2}$ such that if $k,s\geq N_{2}$, then $$\left\|x_{n_{k}}-x_{n_{s}}\right\|<\frac{\varepsilon}{2}.$$ Therefore, considering $N=\max\left\{N_{1},N_{2}\right\}$, if $k,s\geq N$ then $$\left\|g_{n_{k}}-g_{n_{s}}\right\|=\left\|x_{n_{k}}-\lambda_{n_{k}}-x_{n_{s}}+\lambda_{n_{s}}\right\|\leq \left\|x_{n_{k}}-x_{n_{s}}\right\|+\left|\lambda_{n_{k}}-\lambda_{n_{s}}\right|\left\|v\right\|<\varepsilon.$$ So, $(g_{n_{k}})_{k\in\mathbb{N}}\subseteq G$ is Cauchy, then, by the fact mentioned at the beginning, we have that $(g_{n_{k}})_{k\in\mathbb{N}}$ is convergent in $G$.

Therefore, $(x_{n_{k}})_{k\in\mathbb{N}}=(g_{n_{k}}+\lambda_{n_{k}}z)_{k\in\mathbb{N}}$ converges (in $G+F$) because it is sum of two convergent sequences. Thus, $(x_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence with a convergent subsequence in $G+F$, then $(x_{n})_{n\in\mathbb{N}}$ is convergent in $G+F$. Therefore, $G+F$ is closed.
1. $(\lambda_{n})_{nin\mathbb{N}}$ is unbounded. Then there exists a subsequence $(\lambda_{n_{k}})_{k\in\mathbb{N}}$ with $\lim_{k\rightarrow\infty}|\lambda_{n_{k}}|=\infty$. Since $(x_{n})_{nin\mathbb{N}}$ is bounded (by be Cauchy), it follows that $$\left\|z-\frac{1}{\lambda_{n_{k}}}g_{n_{k}}\right\|=\left\|\frac{1}{\lambda_{n_{k}}}x_{n_{k}}\right\|\rightarrow 0,\qquad n\rightarrow \infty.$$ Then $z$ is a limit point of $G$, as $G$ is closed, we have $z\in G$, thus, $G+F=G$. Therefore, $G+F$ is closed.