On the space $X=C[0,1]$, define a norm $||| f |||^2=\Vert f \Vert_{\infty}^2 + \Vert f \Vert_2^2$, where $\Vert . \Vert_\infty$ is the sup norm on $C[0,1]$ space and $\Vert . \Vert_2$ is the $L_2$ norm. This norm is equivalent to that of $\Vert . \Vert_\infty$ on $C[0,1]$ space and $(X, ||| . |||)$ is strictly convex norm. I am trying to show whether $(X, ||| . |||)$ is weakly locally uniformly convex or not. 

My approach: I took $f, f_n$ in $S_{|||. |||}$ 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.