Show that $C^1([0,1])$ is not reflexive

Question

Aim of this exercise is proving that $(C^1([0,1]),\|\cdot\|_{C^1})$ is not reflexive.

We know that, if $(f_h)_h\subset C^1([0,1])$ is a sequence that weakly converges to $f\in C^1([0,1])$ (that is $f_h \rightharpoonup f$),then $(f_h)_h$ and $(f'_h)_h$ pointwise converge to, respectively, $f$ and $f'$ $(*)$.

Now, let $f_h(x)=\frac{x^h}{h}$, for $x\in[0,1]$. From $(*)$ it follows that $f_h(x)$ does not weakly converges in $C^1([0,1])$, because $f'_h=x^{h-1}$ pointwise converges to a discontinuous function.

From this we want to conclude that $(C^1([0,1]),\|\cdot\|_{C^1})$ is not reflexive arguing by contraposition. So, suppose that $(C^1([0,1]),\|\cdot\|_{C^1})$ is reflexive. Then the unit closed ball $B$ is sequentially weakly compact. Now, to reach the absurd, I guess I have to use the sequence $(f_h)$ where $f_h=\frac{x^h}{h}$, but $\|f_h\|_{C^1}=1/h+1>1$. So, what I have to do?

Answer 1

Notice that $\lVert f_h\rVert_{C^1}\leqslant 2$ if we consider $h\geqslant 1$, hence the sequence $(f_h/2)_{h\geqslant 1}$ has all its terms in the closed unit ball of $C^1([0,1])$.