If $E$ is not complemented in $X$, is $E \oplus \{0\}$ not complemented in $X \oplus Y$?

Question

Question: Let $X$ be a Banach space, and let $E \subset X$ be a closed subspace such that $E$ is not complemented in $X$. Does it follow that $E \oplus \{0\}$ is not complemented in $X \oplus Y$, where $Y$ is some other fixed arbitrary Banach space?

I ask this question because I want to understand the answer to this question. I would just ask the answerer for clarification, but it appears that they no longer use this site.

Discussion: Suppose one wants to prove the answer to the question is "yes". A natural approach would seem to be to prove the contrapositive. So, suppose that $F \subset X \oplus Y$ is a closed subspace complementary to $E \oplus \{0\}$. From here we want to argue that $E$ is complemented in $X$. A natural candidate for a complementary subspace to $E$ would be $\pi_1(F)$, where $\pi_1$ is the factor projection $X \oplus Y \to X$. It's simple to see that $\pi_1(F) + E = X$, that is no problem. However:

1. I don't see any reason why $\pi_1(F)$ should be closed in $X$. For example, if $F \subset X \times Y$ is the graph of a closed operator $X \to Y$, then $\pi_1(F)$ is the domain of the operator which needn't be closed.
2. I don't see why $\pi_1(F) \cap E = \{0\}$ should hold.

Thanks for reading.

Answer 1

The answer to my question is "yes". I was just looking at it from the wrong angle somehow. Bit stunned of me, in hindsight.

In fact something slightly more general is true (and basically trivial to prove).

Proposition 1: Let $X \subset Y \subset Z$ where $Z$ is a Banach space and $X,Y$ are closed subspaces. If $X$ is complemented in $Z$, then $X$ is complemented in $Y$ as well.

Proof: Suppose $E$ is a complement for $X$ in $Z$. Then, $F = E \cap Y$is a complement for $X$ in $Y$. Indeed:

• $F$ is closed in $Y$, since $E$ is closed in $Z$.
• Clearly $F \cap X = \{0\}$.
• $F + X \subset Y$, since $F,X \subset Y$.
• Given any $y \in Y$, write $y = e + x$ where $e \in E$, $x \in X$. Note $e = y - x \in Y$, since $y \in Y, x \in X \subset Y$, so that $e \in F$. Thus, $F + X = Y$.

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Added: I had may as well point out that there is statement can be made about quotients.

Proposition 2: As above, suppose there are nested Banach space $X \subset Y \subset Z$. If $Y$ is complemented in $Z$, then $Y/X$ is complemented in $Z/X$.

Briefly, if $E$ is a complement for $Y$ in $Z$, then $E/X$ is a complement for $Y/X$ in $Z/X$.

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Added: Note you can also slightly extend the applicability of Proposition 1 as follows:

Proposition 3: Suppose that $X$ and $Y$ are closed subspaces of a Banach space $Z$. If $X$ is complemented in $Z$, then $X \cap Y$ is complemented in $Y$.

Along similar lines, if $E$ is a complement for $X$ in $Z$, then $E \cap Y$ is a complement for $X \cap Y$ in $Y$.