Is $C^1([a,b], \mathbb{R}^n)$ a reflexive Banach space?

Question

I want to prove or disprove that $C^1([a,b], \mathbb{R}^n)$ equipped with the norm $||x||=\underset{t\in[a,b]}{\sup}|x(t)|_{\mathbb{R}^n}+\underset{t\in[a,b]}{\sup}|\dot{x}(t)|_{\mathbb{R}^n}$ is a reflexive Banach space.

I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space $X$ is reflexive if the mapping $F: X \rightarrow X''$ is surjective, where $X''$ is the second dual space of $X$.

However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.

Is it indeed true that this space is not reflexive?

Answer 1

There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.

It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t \in [a,b]$, consider the linear functional $\ell_t(x) = \dot{x}(t)$. You can check that it is continuous, and that $\|\ell_t - \ell_s\|_{X'} = 2$ for any $s \ne t$. Hence there is an uncountable closed discrete set in $X'$.

You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.

Answer 2

We can embed $Y = C([a,b],\mathbb R)$ into $X = C^1([a,b], \mathbb R^n)$ by $T(f)(x) = \int_a^x f(t)\; dt\; {\bf u}$ where $\bf u$ is some nonzero vector in $\mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.