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I am currently struggling with the following exercise:

Let $B$ be a Banach space and $C, D \subset B$ closed subspaces of $B$.
There is a $M \in ]0, \infty[$ such that $\forall x \in D : \operatorname{dist}(x, C \cap D) \leq M \cdot \operatorname{dist}(x, C)$ holds.

Show that $C + D$ is closed.

bluestool
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D. A.
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2 Answers2

1

Consider the quotient maps $p: B\rightarrow B/C$ and $q: D \rightarrow D/D\cap C$. We claim that $p(D)$ is closed in $B/C$. Indeed if $\{p(d_n)\}_n$ is Cauchy in $B/C$ then by your assumption on the distances $\{q(d_n)\}_n$ is Cauchy in $D/D\cap C$ and since this space is complete $\{q(d_n)\}_n$ converges to a $q(d)$.Then it is easy to see that $p(d_n)\rightarrow p(d)$. Having verified that $p(D)$ is closed, it follows that $p^{-1}(p(D))=D+C$ is closed.

Jonathan Hole
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I have proven it as follows:

$B$ is a Banach space and $D, C \subseteq B$ are closed. Thus $D$ and $C$ are themselves Banach spaces. $C \cap D$ is also a closed subspace of B and thus $D / C \cap D$ is a Banach space.

The second isomorphism theorem states that $D / C \cap D \simeq (D + C)/D$ given by the isomorphism of vector spaces $\varphi : D / C \cap D \to (D + C)/C, D + C \cap D \mapsto D + C$.

We now show that $\varphi$ and $\varphi^{-1}$ are continuous:

We know that $C \cap D \subseteq D \implies \forall x \in D : ||[x]||_{(D+C)/C} = \operatorname{dist}(x, C) \leq \operatorname{dist}(x, C \cap D) = ||[x]||_{D/C \cap D}$. Thus $\varphi$ is continuous.

Now show that $\varphi^{-1} : (D+C)/C \to D / C \cap D, [D+C] \mapsto [D]$ is continuous: The inequation given results in: $\forall D \in D, C \in C : ||[D]||_{D / C \cap D} = \operatorname{dist}(D, C \cap D) \leq M \cdot \operatorname{dist}(D, C) = M \cdot ||[D]||_{(D+C)/C}$ which proves that $\varphi^{-1}$ is continuous.

Using $\varphi$ we see that $(D+C)/C$ is a Banach space, since $D / C \cap D$ is one.

Let $(x_n)_{n \in \mathbb N}$ be a Cauchy sequence. $(x_n)$ converges to a $z \in B$, since $B$ is complete. Now look at the induced Cauchy sequence $([x_n])_{n \in \mathbb N}$. It converges in the Banach space $(D + C)/C$ to $[z] \in (D + C)/C$.

Thus we have $||[z] - [x_n]||_{(D+C)/C} = \operatorname{dist}(z - x_n, C) \overset{n \to \infty}{\longrightarrow} 0$. Since $C$ is closed, we also have $\operatorname{dist}(z - x_n, C) \overset{n \to \infty}{\longrightarrow} \operatorname{dist}(z - x, C)$.

Thus it follows that $\operatorname{dist}(z - x, C) = 0 \overset{C~\text{closed}}{\implies} z -x \in C \implies z - x \in C \implies x \in z + C \subseteq D + C$.

Thus we have $\overline{D+C} \subseteq D + C$ which shows that $D + C$ is closed. qed.

D. A.
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