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Consider $X =C([-1,1])$ with the usual norm $\lVert f \rVert_{\infty} = \sup_{t\in [-1,1]} |f(t)|.$ Let \begin{align*} A_+ &= \{ f \in X: f(t) = f(-t) \quad \forall t \in[-1,1] \}\\ A_- &= \{ f \in X: f(t) = -f(-t) \quad \forall t\in [-1,1] \}. \end{align*} Show that $A_+$ and $A_{-}$ are meager.

I'm uncertain on how to proceed with this.

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    $\begingroup$ You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed? $\endgroup$ – Mathematician 42 Dec 9 '17 at 21:03
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    $\begingroup$ As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior. $\endgroup$ – Martin Sleziak Dec 10 '17 at 11:20
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For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.

First, we note what an $\varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $\varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $\varepsilon$ “tube” around $f$.

To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g \in U$, then $\exists T \in [0,1]$ such that $g(-T) \neq g(T)$. Define $\delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $\frac{\delta}{3}$ ball around $g$, it’s clear that $h(T) \neq h(-T)$. In other words, the entire $\frac{\delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.

To show its interior is empty, pick any $f\in A_{+}$ and consider an $\varepsilon$-ball $B$ around $f$. Since this is an $\varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $s\in B$ such that for some $T, s(T) \neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.

The arguments are analogous for $A_{-}$.

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    $\begingroup$ Seems easier to me to use sequences to show $A+$ is closed. $\endgroup$ – zhw. Dec 9 '17 at 23:47

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