Suppose $X,Y$ are normed spaces and $T:X\to Y$ is bounded. Suppose further that $\exists r>0$ such that $\|T^{*}f\|\ge r\|f\| \;\;\;\forall f\in Y'$ where $T^{*}:Y'\to X'$ is given by $T^{*}(f)=f\circ T$. Show that $B_Y(r)$, the closed ball of radius $r$ centered at $0$ is contained in $C:=\overline{T(B_X)}$ where $B_X$ is closed unit ball in $X$.
I think this is about HST ( Hahn Banach Separation theorem).
Attempt: Suppose that there is $y_0\in B_Y(r)\setminus C$. Then separation theorem states that there is $f \in Y'$ such that
$$ \Re f(y_0) > \sup_{c\in C} \Re f(c) $$
But also
$$ \Re f(y_0) \le |f(y_0)| \le \|f\| \|y_0\| \le \|f\| r \le \|T^{*} f\| $$
and I'm stuck. I wanted to somehow show contradiction, but it seems I'm not going anywhere with my plan.
It'd be nice if someone can show how to do this!
To arrive to a contradiction, you also need to compute $\|T^*f\|$:
$$\|T^*f\|=\sup\{T^*f(x): x\in B_x\}= \sup \{ f(Tx): x\in B_x\} \leq \sup f(\overline{T(B_X)}).$$ Then you combine it with the two inequalities that you already wrote.