On the space $X=C[0,1]$, define a norm $||| f |||^2=\Vert f \Vert_{\infty}^2 + \Vert f \Vert_2^2$, where $\Vert \cdot \Vert_\infty$ is the sup norm on $C[0,1]$ space and $\Vert \cdot \Vert_2$ is the $L_2$ norm. This norm is equivalent to that of $\Vert \cdot \Vert_\infty$ on $C[0,1]$ space and $(X, ||| \cdot |||)$ is strictly convex norm. I am trying to show whether $(X, ||| \cdot |||)$ is weakly locally uniformly convex or not.
My approach: I took $f, f_n$ in $S_{|||\cdot |||}$ with $||| f + f_n ||| \to 2$. Upon solving, I got $\Vert f \Vert_\infty \Vert f_n \Vert_\infty + \int_0^1 f(t)f_n(t) \, dt \to 1.$ I do not know any further step to conclude that $f_n \to f$ weakly. Kindly help me. Thank you in advance.
Best Answer
Your norm, call it $N$ for \TeX-nical simplicity, is not WLUR. You want to know whether $N(f)=1$, $N(f_n)\to1$, $N(f+f_n)\to 2$ imply that $f_n\to f$ weakly. For a counterexample let $f$ be the constant function $f=1/\sqrt2$ and $f_n(x)=1/\sqrt2$ for $x\ge 1/n$, $f_n(0)=0$ and $f_n$ linear in between. (Another choice would be $f_n(x)= (1-x^n)/\sqrt2$ if one prefers polynomials.) Weak convergence fails since $f_n(0) \not\to f(0)$.