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$C(I_n)$ is the space of continuous functions on $I_n=[0,1]^n$ with the metric $d(f,g)=\sup\{|f(x)-g(x)\mid x\in I_d\}$ where $f,g\in C(I_n)$. M is a linear subspace of $C(I_n)$.

If $f_0\in C(I_n)\setminus \operatorname{cl}(M)$, then there is a $\delta>0$ for which $\sup|f(x)-f_0(x)|>\delta\enspace\forall f \in M$

My question is, if the closure of $M$, $\operatorname{cl}(M)$, is a closed proper subset of $C(I_n)$, does this imply that $\sup|f(x)-f_0(x)|>\delta\enspace\forall f \in \operatorname{cl}(M)$ also?

Bernard
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Stokastisk
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1 Answers1

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Yes.

You know that if $f_0 \notin \overline{M}$ then $d(f_0, M) = \delta > 0$, where $d(f_0, M) = \inf_{f \in M} d(f_0, f)$.

By the same reasoning, we have $\overline{\overline{M}} = \overline{M}$ so if $f_0 \notin \overline{\overline{M}}$ then $d(f_0, \overline{M}) = \delta > 0$.

Also in general it holds $d(f_0, M) = d(f_0, \overline{M})$, as you can see here.

mechanodroid
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