I know $(C^1([a,b], \left\lVert\cdot\right\rVert_{C^1})$ is not reflexive (and neither $(C^0([a,b]),\left\lVert \cdot\right\rVert_{\infty})$ is), but I was wondering if maybe using a weaker norm we can make it reflexive (altough I don't think it is true).
I know that if $E$ is a reflexive space then the closed unit ball is weakly sequentially compact, and $B_{(C^1([a,b]),\left\lVert \cdot\right\rVert_{\infty})}\subset B_{(C^0([a,b]),\left\lVert \cdot\right\rVert_{\infty})}$ which is not weakly sequentially compact, but I don't know how this would help me to conclude.
Moreover, I suspect $(C^1([a,b]),\left\lVert \cdot\right\rVert_{\infty})$ is not closed in $(C^0([a,b]),\left\lVert \cdot\right\rVert_{\infty})$ (take $f_n(x)=|x|^{1+\frac{1}{n}}$: $f_n$'s are continuous but they're derivatives are not), but again, how closedness could help me is a mistery.
Any hint would be appreciate, thanks in advance.
Best Answer
A non-reflexive Banach space $(X,\|\cdot\|)$ can't have a weaker norm $|\cdot |$ making it reflexive, because reflexive normed spaces are complete (the dual of any normed space is complete for the dual norm) and thus, Banach's isomorphy theorem (the open mapping theorem) implies that both norms $\|\cdot\|$ and $|\cdot|$ are equivalent so that $(X,\|\cdot\|)$ would be reflexive.