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

Solution 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 $x \in F+G$.

Solution 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.