Norms and pointwise convergence It is known that


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*There is no norm $\| \cdot\|$ on the space $E$ of continuous real-valued functions on an interval, say $[0,1]$ such that $f_n \to f$ for $\|\cdot\|$ if and only if $f_n$ converges pointwise to $f$, cf Norm for pointwise convergence

*For any finite dimensional subspace $F$ of $E$ there is such a norm (for polynomials, on can use the Lagrange interpolating polynomials).
Hence my question: is there an infinite dimensional subspace of $E$ having such a norm.
Note that the problem concerns only sequences: it is not about inducing the pointwise convergence topology
Thanks
 A: $\;\;\;\;$ In the range $\text{Im}(\Psi):=\{f\in\mathbf{C^*[0,1)}:\lim_{x\to 1}\;f(x)\;\text{exists}\}$ of the $\mathbf{C[0,1]}$ restriction map $\Psi:\;f\mapsto f|_{[0,1)}$ we carve out an example of an infinite dimensional closed subspace $\mathcal U$ in which pointwise and normwise $(\Vert\cdot\Vert_\infty)$ convergence are proven to be one in the same, which suffices because $\mathcal U$ and $\text{Im}(\Psi)$ are embedded in $\mathbf{C[0,1]}$ via the isometry ring isomorphism $\Psi^{-1}$.
$\;\;\;\;$ Fix a strictly monotonic sequence $(x_k)_{k\geq 0}$ with $x_0=0$ and $\lim_{k\to\infty}\;x_k=1$. Let $\mathcal U$ be the space of all piecewise linear functions $f$ with the following two properties :
$(\mathbf I)$ $f$ is piecewise linear with domain partition $\mathcal A:= \{I_k : k\geq 0\}$ where $I_k:=[x_k,x_{k+1})$
$(\mathbf {II})$ $\forall k\geq 0\;|f^{(k)}|\geq 2|f^{(k+1)}|$ where $f^{(k)}$ denotes the constant value of $f'(x)$ on $(x_k,x_{k+1})$.
$\;\;\;\;$ As always, normwise convergence implies pointwise convergence but now assume the $\mathcal U$ sequence $(f_n)_{n\geq 0}$ converges pointwise to $F$. $F|_{I_k}$ is always linear as it is both the pointwise and normwise $(||\cdot||_\infty)$ function limit of the linear function sequence $(f_n|_{I_k})_{n\geq 0}$. $F$ is continuous as $\lim_{x\to {x_{k+1}}}f_n|_{I_k}(x)=f_n|_{I_{k+1}}(x_{k+1})$ gaurantees  $\lim_{x\to {x_{k+1}}}F|_{I_k}(x)=F|_{I_{k+1}}(x_{k+1})$ implying continuity at each $x_k$.
$\;\;\;\;$ Therefore, $F$ satisfies property $\mathbf{(I)}$. $F$ must satisfy property $\mathbf{(II)}$, i.e. $|F^{(k)}|\geq 2|F^{(k+1)}|$, because $|f_n^{(k)}|\geq 2|f_n^{(k+1)}|$. Therefore, $F\in\mathcal U$. Therefore, $\mathcal U$ must be closed pointwise.
$\;\;\;\;$ Finally, it suffices to show $f_n\to F$ normwise. Fix $\epsilon\in (0,1)$. Fix $K\in\Bbb N$ such that both $1-x_K<\frac{\epsilon}{2}$ and $|f_n^{(K)}|,|F^{(K)}|<\frac{1}{2}$.
Fix $N\in\Bbb N$ such that $\;\;n>N\implies\forall k\leq K\;\;|f_n(x_k)-F(x_k)|<\frac{\epsilon}{2}$ $$\text{i.e.}\;\;\;\;\; n>N\implies\Vert f_n|_{[0,x_K]}-F|_{[0,x_K]}\Vert_\infty<\frac{\epsilon}{2}$$
Since $|f_n'(x)-F'(x)|\leq 1$ on $(x_K,1)$, except at $x=x_k\;\;(k>K)$, we must have 
$$n>N\implies\Vert f_n|_{[x_K,1)}-F|_{[x_K,1)}\Vert_\infty\leq |f_n(x_K)-F(x_K)|+(1-x_K)<\frac{\epsilon}{2}+\frac{\epsilon}{2}$$
$$\therefore\;n>N\implies\Vert f_n-F\Vert_\infty<\epsilon$$
$$\therefore\;f_n\to F\;\;\text{normwise}$$ 
