I have tried to resolve this exercise but I am not sure about how to do it.
Let $c_0$ be the subespace of $l_\infty$ formed by the convergent to 0 sequences
a) Find all the elements with norm 1 of the dual of $c_0$, $c_0^*$, such that they reach the norm for $y_0=(1, 1, 1/2, 1/3,...)$.
b) For the previous elements, find all the vectors in the unit closed ball of $c_0$ where those elements reach their norm.
c) Is $c_0$ strictly convex?
Since I don't know the answer for a) I can't move forward to the next part.
a): You have to find all $(a_n) \in L^{1}$ such that $\sum |a_n|=1$ and $a_1+a_2+a_3/2+a_4/3+...=1$. This gives $1=a_1+a_2+a_3/2+a_4/3+... \leq \sum |a_n|=1$ so equality must hold throughout. This is possible only when $a_2=a_3=....=0$, and $0\leq a_1 \leq 1, a_2=1-a_1$. Hence the answer to a) is $\{(a,1-a,0,0...): 0\leq a \leq 1$.
b): You have to find all sequences $(b_n) \in c_0$ such that $|b_n| \leq 1$ for all $n$ and $ab_1+(1-a)b_2=\|(a,1-a,0,..)\|=1$. I will let you show that this is possibel only whe $b-1=b_2=1$.
c) Since $\|(1,1,0,0...)\|=1$, $\|(1,-1,0,0...)\|=1$ and $ \|(1,1,0,0...) +(1,-1,0,0...)\|=2$ the space is not strictly convex.