Let $X$ be a Banach space and let $Y=BC(\mathbb{R},X)$ be the Banach space of all bounded continuous functions from $\mathbb{R}$ to $X$ equipped with the supremum norm. Let $(f_n)_n$ be a sequence of functions, $f_n\in Y$ and $f\in Y$. Is there any link between the following convergences:

$$\sup_{t\in\mathbb{R}} |\langle x^*,f_n(t)-f(t)\rangle|\to 0,$$ for all $x^*\in X^*$ as $n\to\infty$.

And the convergence $$\langle \varphi^*,f_n-f\rangle\to0$$ for all $\varphi^*\in Y^*$.

$X^*$ and $Y^*$ denote the dual spaces of $X$ and $Y$.


1 Answer 1


If I haven't made mistakes, none of the implications hold. Let $$\begin{split}(A)&\qquad\sup_{t\in\mathbb{R}} |\langle x^*,f_n(t)-f(t)\rangle|\to 0,\\(B)&\qquad\langle \varphi^*,f_n-f\rangle\to0.\end{split}$$

$(B)\Rightarrow (A)$ fails even when $X=\mathbb{R}$: take $$g(x):=\begin{cases} 0 &\text{ if }x\le 0 \\ 2x &\text{ if }0\le x\le\frac{1}{2} \\ 2-2x &\text{ if }\frac{1}{2}\le x\le 1 \\ 0 &\text{ if }x\ge 1 \end{cases}$$ and set $f_n(x):=g(nx)$. Then $f_n(x)\to 0$ pointwise but not uniformly, so $(A)$ does not hold.
But $BC(\mathbb{R},\mathbb{R})=C(\beta\mathbb{R},\mathbb{R})$ ($\beta\mathbb{R}$ is the Stone-Cech compactification of $\mathbb{R}$: any function in $BC(\mathbb{R},\mathbb{R})$ extends uniquely to $\beta\mathbb{R}$ by the universal property), so you can see any $\phi^*\in Y^*$ as a Radon (signed) measure on $\beta\mathbb{R}$.
Now the extensions $\tilde{f_n}:\beta\mathbb{R}\to\mathbb{R}$ still converge pointwise to $0$ (as their support is still contained in $[0,1]$, tell me if this is not clear). So by dominated convergence $\langle\phi^*,f_n\rangle\to 0$, i.e. $(B)$ holds.

$(A)\Rightarrow (B)$ is true when $X$ is finite-dimensional (see comments), but fails when $X=\ell^2(\mathbb{N})$: let $P_n:X\to X$ be the orthogonal projection onto $\langle e_1,\dots,e_n\rangle$, so that for any $x$ we have $P_n x\to x$.
Let $V\subset Y$ be the subspace of functions whose range is contained in a finite dimensional subspace of $X$. Now take $$f(x):=\begin{cases} 0 &\text{ if }x\le 0 \\ e_n &\text{ if }x=n \end{cases}$$ (and define $f$ elsewhere in a piecewise linear fashion). Take $f_n:=P_n\circ f$. Now $\langle x,f_n(t)-f(t)\rangle=\langle P_n x-x,f(t)\rangle\to 0$ uniformly as $n\to\infty$ for any $x\in X$, so $(A)$ holds.
Now it suffices to see that $f\not\in\overline{V}$: then by Hahn-Banach there exists $\phi^*\in Y^*$ vanishing on $V$ and s.t. $\langle \phi^*,f\rangle=1$ (so $\langle \phi^*,f_n\rangle=0$ and $(B)$ fails).
Suppose by contradiction $f\in\overline{V}$. Then there exists $g\in Y$ s.t. $\|f-g\|_\infty<\frac{1}{10}$. Let $E\subset X$ be a finite dimensional subspace s.t. $g(\mathbb{R})\subseteq E$. As $g$ is bounded, $g(\mathbb{R})\subseteq K$ for some compact $K\subset E$. Now choose finitely many points $x_1,\dots,x_N\in K$ such that for any $x\in K$ we have $\|x-x_i\|<\frac{1}{2}$ for some $i$. Let $m\in\mathbb{N}$ be so big that $|\langle x_i,e_m\rangle|<\frac{1}{4}$ for any $i$. Now $\|x_i-e_m\|^2>1+\|x_i\|^2-2\cdot\frac{1}{4}\ge\frac{1}{2}$, so for any $x\in K$ it holds $\|x-e_m\|>\frac{1}{\sqrt{2}}-\frac{1}{2}>\frac{1}{10}$, contradicting $d(f(n),K)<\frac{1}{10}$.


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