pointwise convergence on $S \Leftrightarrow$ uniform convergence on $[0,1]$

Question

Let $S \subset [0,1], \mid S \mid < \infty$ ($S$ is finite) and $F$ a sub-vector space of $C^0([0,1], \mathbb{R})$.

Then find a sufficient and necessary condition on $F$ such that pointwise convergence on $S$ is equivalent to uniform convergence on $[0,1]$.

What I've noticed so far :

For all element of $F$ we must have : $\exists s \in S, f(s) \ne0$ otherwise it's possible we don't even have pointwise convergence.

Now, the problem is that I need to have uniform convergence and not only pointwise convergence... But I am enable to find a sufficient and necessary condition from here

Answer 1

Proposition. The following conditions are equivalent.

1. For each sequence $\{f_n\}$ of functions of $F$ and each function $f\in F$, $\{f_n\}$ converges to $f$ uniformly on $[0,1]$ iff $\{f_n|S\}$ converges to $f|S$ pointwise on $S$.
2. For each function $f\in F$ the restriction map $F\to C^0(S,\Bbb R)$, $f\mapsto f|S$ is injective
3. For each $f\in F$, if $f|S$ is the zero function $0_S$ then $f$ is the zero function $0_{[0,1]}$.
   Remark that if any of these conditions holds then the dimension of $F$ is at most $|S|$.

Proof. The equivalence of 2 and 3 is a simple fact about homomorphisms between vector spaces.

$1\Rightarrow 3$. Let $f\in F$ be a function such that $f|S=0_S$. Then a sequence $\{f|S, 0_{[0,1]}|S,f|S, 0_{[0,1]}|S,\dots \}$ converges pointwise on $S$, so a sequence $\{f, 0_{[0,1]},f, 0_{[0,1]},\dots \}$ converges uniformly on $[0,1]$. This is possible only if $f=0_{[0,1]}$.

$3\Rightarrow 1$. Let $\{f_n\}$ be a sequence of functions of $F$ and $f\in F$. If the sequence $\{f_n\}$ converges to $f$ uniformly on $[0,1]$, then a sequence $\{f_n|S\}$ converges to $f|S$ pointwise on $S$.

Conversely, assume that the sequence $\{f_n|S\}$ converges to $f|S$ pointwise on $S$. Let $B=\{b_1,\dots, b_k\}$ be a basis of the space $F$. For each $n$ let $f_n=x^n_1b_1+\dots+x^n_kb_k$ be the decomposition of $f_n$ with respect to the basis $B$.

We claim that for each $i=1,\dots, k$ a sequence $\{x^n_i\}$ is fundamental. To assure this it sufffices to show that there exists $c_i>0$ such that for each $g=y_1b_1+\dots+y_kb_k\in F$, we have $\|g\|_S=\max \{|g(s)|:s\in S\}\ge c_i|y_i|$. Condition 3 assures that a vector $b_i|S$ does not belong to a linear hull $L_i$ spanned in $\Bbb R^S$ by all vectors $b_j|S $ with $j\ne i$. Since $L_i$ is a closed subset of $\Bbb R^S$ (we can prove this by induction with respect to $k$), we can put $$c_i=\inf \{\| b_i|S-v\|_S:v\in L_i\}>0.$$

Therefore there exists $x_i=\lim_{n\to\infty} x^n_i$. Put $g=x_1b_1+\dots+x_kb_k\in F$. Since each $b_i$ is a continuous function on a compact set $[0,1]$, it is bounded. Put $M=\sup\{b_i(t):i=1,\dots, k; t\in [0,1]\}$. Let $\varepsilon>0$ be an arbirary number. There exists number $N$ such that $|x^n_i-x_i|<\varepsilon$ for each $n>N$ and each $i=1,\dots, k$. For each $h\in C^0([0,1],\Bbb R)$ put $\|h\|=\sup\{|h(t)|:t\in [0,1]\}$.

Then

$$\|f_n-g\|=\| x^n_1b_1+\dots+x^n_kb_k - x_1b_1+\dots+x_kb_k \|\le |x^n_1-x_1|\|b_1\|+\dots+|x^n_k-x_k|\|b_1\|\le k\varepsilon M.$$

That is, the sequence $\{f_n \}$ converges to $g$ uniformly on $[0,1]$. Since the sequence $\{f_n|S\}$ converges to $f|S$ pointwise on $S$, $g|S=f|S$. Therefore $g=f$. $\square$