Analysis Problem of the Day 68

Today’s problem appeared as Problem 8 on the Texas A&M Winter 2023 Real Analysis Qual:

Problem 8. a) Let $X$ be a normed vector space and let $Y$ be a subspace of $X.$ Show that if $Y$ has nonempty interior, then $Y=X.$

b) Let $X$ be a Banach space and $T:X \to X$ be a bounded operator. Show that if for each $x \in X,$ $T^n x=0,$ then there exists $d$ such that $T^d =0.$

Solution: a) Since $Y$ has nonempty interior, for some $\epsilon>0,$ $B(y,\epsilon) \subset Y.$ In particular, $y \in Y,$ so since $Y$ is a subspace, $-y \in Y.$ Thus, $B(y,\epsilon)-y = B(0,\epsilon) \subset Y.$ But this implies that for any $x \in X, \frac{\epsilon}{2}\frac{x}{\|x\|} \in Y,$ and since $Y$ is a subspace, $x \in Y.$ Thus, $X=Y.$

b) Notice that $X$ is a Banach space. This means that completeness is important. In particular, if we define $A_n :=\ker (T^n),$ then since $T^n$ is a bounded (and therefore continuous) operator, $A_n$ are closed subspaces of $X,$ and by assumption, $\bigcup_{n=1}^\infty A_n = X.$ By the Baire category theorem (which is what requires completeness), it follows that some $A_d$ has nonempty interior, and since $A_d$ is a subspace, by (a) one has $A_d=X,$ i.e. $T^d =0$ for some $d \in \mathbb{N}.$

Remark: In general, if one has some pointwise property in a complete metric space, one may upgrade it to a uniform property using the Baire category theorem. For example, the theorem is used directly in the proof of the Open Mapping Theorem, Closed Graph Theorem, and Uniform Boundedness Principle.

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