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Let $[a,b]$ be an interval and $X=C([a,b])$ be the vector space of all continuous real-valued functions on $[a,b]$ equipped with the norm $||f||=\max_{a \leq t \leq b}|f(t)|$. Let $F$ be a fixed continuous function such that $F(t)>0$ for all $t \in [a,b]$ and let $$S=\{f \in X : |f(t)|<F(t) \operatorname{for all} t \in [a,b]\}$$

Prove that S is open, find $\operatorname{cl}(S)$ (closure), and find $\partial S$ (boundary).

$S \subset X$ is open if for any $x \in S$, there exists $r_x>0$ such that $D(x, r_x) \subset S$. I also know that any open ball in a metric space is an open set, and that $S \subset X$ is open if and only if $S=\cup_{i \in I}D(x_i,r_i)$, where $\{D(x_i,r_i)\}_{i \in I}$ is a family of open balls in $X$. I know that these properties can be used to prove that $S$ is open, but I don't know how.

In regards to closure, I know that if $(X,d)$ is a metric space and $S \subset X$, then $\operatorname{cl}(S)=\{x \in X : \forall \epsilon>0, D(x,\epsilon) \cap S \neq \emptyset\}$. I don't know how to express it in my specific case, however.

Lastly, I know that $\partial S=\{x \in X : \forall \epsilon>0, D(x,\epsilon) \cap S \neq \emptyset\}\cap \{x \in X : \forall \epsilon>0, D(x,\epsilon) \cap S^c \neq \emptyset\}=\operatorname{cl}(S) \cap \operatorname{cl}(S^c)$, where $S^c$ is the complement of $S$, and also that $\partial S \subset \operatorname{cl}(S)$.

Any assistance with these three questions would be greatly appreciated. Thank you!

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Let $f\in S$, and consider $$\lambda_f:=\min_{x\in [a,b]} F(x)-|f(x)|>0$$ The open ball $B(f,\lambda_f/2)$ is contained in $S$: infact, given $g\in B(f,\lambda_f/2)$, we have \begin{align*} \min_{x\in [a,b]}F(x)-|g(x)|&=\min_{x\in [a,b]}F(x)-|g(x)|+|f(x)|-|f(x)|\\ &\geq\min_{x\in [a,b]} F(x)-|f(x)|-\max_{x\in [a,b]}|f(x)-g(x)|\\ &\geq \lambda_f-\lambda_f/2>0 \end{align*} Hence $S$ is open. Now let's show that $$cl(S)=\{f\in X:|f(x)|\leq F(x)\,\forall x\in[a,b]\}$$ Denote $T:=\{f\in X:|f(x)|\leq F(x)\,\forall x\in[a,b]\}$. We prove firstly that $T$ is closed. We are working in the metric space $X$, so it sufficies that $T$ contains all its limit points. Let $f_n\in T$, $f_n\rightarrow f$ (with respect to the sup norm). In particular, for all $x\in[a,b]$, $|f_n(x)|\rightarrow |f(x)|$. Moreover, for all $n$, we have $|f_n(x)|\leq F(x)$. Passing to the limit for $n\rightarrow\infty$, we obtain that for all $x\in [a,b]$ $$|f(x)|\leq \sup_n|f_n(x)|\leq F(x)$$ Then $f$ belongs to $T$. To conclude that $cl(S)=T$, we only need to prove that given $f\in T$, it's actually a limit point for $S$. Infact, by setting $f_n:= (1-1/n)f$, we have $ |f_n(x)|\leq (1-1/n)F(x)<F(x) $. Hence $f_n\in S$ for all $n$, and $f_n\rightarrow f$, so $f$ is a limit point for $S$. For the bounfary, just observe that $$\partial S=cl(S)\backslash S=\{f\in T:|f(x)|=F(x)\mbox{ for some }x\in [a,b]\}$$

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  • $\begingroup$ So, does this imply that $\operatorname{cl}(S)=S$, in the above question? Is this a common occurrence? $\endgroup$ – Douglas Fir Feb 13 '15 at 2:51
  • $\begingroup$ No, the closure of $S$ is a strictly large set. Indeed the boundary of $S$ is not empty $\endgroup$ – Capublanca Feb 14 '15 at 15:36
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Hint:

To show $S$ is open, let $f\in S$. Then you want to find $\epsilon >0$ so that the ball $D(f, \epsilon) \subset S$. As $f\in S$, by definition $|f(t)| < F(t)$ for all $t \in [a, b]$. As $F, f$ are both continuous, there is $\epsilon >0$ so that

$$ |f(t) | + \epsilon < F(t)$$

for all $t\in [a, b]$.

To find the closure, let $g\in cl(S)$. Then for all $\epsilon >0$, there is $f \in S$ so that $d(f, g) <\epsilon$. That is

$$|g(t)| \leq |g(t) - f(t)| + |f(t)| \leq \epsilon +F(t)$$

for all $t\in [a, b]$.

To find the boundary, note that $\partial S = cl(S)\setminus S$ (as $S$ is open).

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