uniform continuity of the function $t\mapsto\langle x^*,f(t)\rangle$

Question

Let $X$ be a Banach space. $f:\mathbb{R}\to X$ a function. If we have $t\mapsto\langle x^*,f(t)\rangle$ uniformly continuous on $\mathbb{R}$ for each $x^*\in D$ where $D$ is a dense subset of $X^*$ (dual space of $X$). Do we have $t\mapsto\langle x^*,f(t)\rangle$ uniformly continuous on $\mathbb{R}$ for each $x^*\in X^*$. If not can we add an additional assumption to make this true ?

Answer 1

No, the passage from a dense subset to everything is not always possible. For example, let $X=\ell^2$ with standard basis $(e_n)$ and define $$ f(t) = \sum_{n=2}^\infty n(1-n|t-n|)^+ e_n $$ where as usual, $a^+=\max(0,a)$. This is a continuous function.

Let $D=c_{00}$, the set of eventually zero sequences. For every $y\in D$ the function $t\mapsto \langle f(t),y\rangle $ is uniformly continuous, because it is continuous and compactly supported.

On the other hand, for $y=(1/j: j\ge 1)\in \ell^2$ the function $$ g(t) = \langle f(t),y\rangle = \sum_{n=2}^\infty (1-n|t-n|)^+ $$ is not uniformly continuous because $|g(n)-g(n+1/n)|=1$, $n=2,3,\dots $.

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Answer to a misunderstood question: does weak uniform continuity imply uniform continuity of $f$?

No, even if $D=X^*$. For example, let $X=\ell^2$ with standard basis $(e_n)$ and define $$ f(t) = \sum_{n=2}^\infty (1-n|t-n|)^+ e_n $$ where as usual, $a^+=\max(0,a)$. The function $f$ is not uniformly continuous because $$\|f(n)-f(n+1/n)\|=1,\quad n=2,3,\dots $$ On the other hand, for every $y\in \ell^2$ the function $t\mapsto \langle f(t),y\rangle $ is uniformly continuous, because it is continuous on $\mathbb R$ and $\lim_{t\to\infty} f(t)=0$.

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As Olivier Bégassat said in a comment, having uniformly equicontinuous functions $t\mapsto \langle x^*, f(t)\rangle$ over a dense subset $D$ of the unit ball of $X^*$ is enough. In fact this is a necessary and sufficient condition for uniform continuity of $f$. The necessity is straightforward. So is the sufficiency; given $\epsilon>0$, choose $\delta>0$ as in the definition of uniform equicontinuity. Then if $|t_1-t_2|<\delta$, $$\| f(t_1)-f(t_2)\| = \sup_{x^*\in D}\langle x^*, f(t_1)-f(t_2)\rangle \le \epsilon$$ as desired.