I am currently learning the basics of functional analysis and wanted to know if I understand correctly the basic concepts. Therefore I would be thankful for verifying my proofs of a couple of simple exercises.

Let $X:=C[a,b]$, where $-\infty<a<b<\infty$ and $||u||:=max_{a\leq x \leq b}|u(x)|$. We have to show that:

a) $M:=\{ u\in X: u(a)>0 \}$ is an open, not dense subset of $X$.

b) $M:= \{ u\in X: u(a) = 1 \}$ is a closed, not dense subset of $X$.

c) $M:= \{ u\in X: u(x)=0 \text{ on } [c,d] \}$ is not dense provided $a\leq c < d \leq b$.

Ad. a)

$M$ is open iff $\forall u \in M$ $\exists \epsilon > 0$ s.t $\epsilon$-neighbourhood $U_{\epsilon}(u)$ ($U_{\epsilon}(u):= \{ v\in X: ||v-u||<\epsilon \}$) of $u$ is contained in $M$. If $||v-u||<\epsilon$, then it implies that $$ |v(a)-u(a)|<\epsilon, $$ so if we take $\epsilon>0$ such that $u(a)-\epsilon >0$, then for all $v \in X$ satisfying $||v-u||<\epsilon$ we will have $$ u(a)-\epsilon<v(a)< u(a) + \epsilon \implies 0<u(a)-\epsilon < v(a), $$ so $v \in M$ and thus $U_{\epsilon}(u) \subseteq M$. For the density part observe that $\bar{M} = \{ u\in X: u(a)\geq0 \}$ and hence $\bar{M}\neq X$ (because there are functions $v \in X$ such that $v(a)<0$), so M is not dense.

Ad. b)

Take some sequence $\{ u_{n} \} \subset M$ such that $u_n \to u$ as $n \to \infty$. Then, by the sup norm, $u_n \to u$ uniformly and thus the limiting function $u$ is continuous. Hence we can take the limit and obtain $$ u_{n}(a)=1 \quad \forall n\in\mathbb{N} \implies u(a) = 1, $$ so $u\in M$, which is equivalent to the set $M$ being closed. Then, since $M$ is closed we have $M=\bar{M}$ and of course $\bar{M}\neq X$, so $M$ is not dense.

Ad. c)

Since $M$ is closed we have $M=\bar{M}$ and of course $\bar{M}\neq X$ (since there are continuous functions in $X$ which can take a nonzero values in any given interval $[c,d]$)

I would be very thankful for any feedback as I'm still learning the basics and want to know that I understand them well.

  • $\begingroup$ Looks good. BTW a general result is that if $X, Y$ are metric spaces and a sequence $f_n:X\to Y$ of continuous functions converges UNIFORMLY to $f:X\to Y$ then $f$ is continuous. An immediate corollary is that the metric $d(f,g)=\|f-g\|$ on $C[a,b]$ is a complete metric . $\endgroup$ Apr 2 '17 at 3:08

Your solutions are great. Just two small comments:

  1. For $(a)$, you need to justify why $\overline{M}$ is $\{ u(x) \, | \, u(a) \geq 0 \}$. However, to deduce that $M$ is not dense, it is enough to show the even easier statement that $\overline{M} \subseteq \{ u(x) \, | \, u(a) \geq 0 \}$. This holds because if $u_n \in M$ and $u_n \to u$ then $u_n$ converges to $u$ uniformly and in particular we have pointwise convergence so $u(a) = \lim_{n \to \infty} u_n(a)$ and each $u_n(a) > 0$ so the limit satisfies $u(a) \geq 0$.
  2. For $(c)$, you again need to justify the statement that $M$ is closed. This holds again because uniform convergence implies pointwise convergence.
  • $\begingroup$ That's right. Thanks for pointing out those flaws! $\endgroup$
    – Mat Dyl
    Apr 1 '17 at 22:21

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