Strong closure vs. weak-* closure Real fact
It is a known fact that, if $X$ is Banach and $C\subseteq X$ is a strongly closed convex set, then $C$ is also weakly closed. The proof goes like this:

Pick $x_0\notin C$. $\{x_0\}$ is compact and $C$ closed, they're both convex, so by Hahn-Banach there is $f\in X'$ which separates the two strictly, that is, $f(x)<\alpha<f(x_0)$ for some $\alpha\in\mathbb R$ and all $x\in C$. Then, $V:=f^{-1}([\alpha,\infty))$ is a weakly open set containing $x_0$ and disjoint from $C$, so that $x_0$ isn't in the weak closure of $C$. We just proved $X\smallsetminus C$ is open in the weak topology, which concludes the proof.

Hearsay
I've recently been told such a fact holds for weak-* closure.
Generalization attempt: fail
For the moment, I generalized the above to prove the following.
Lemma
If $X$ is Banach and $C\subseteq X'$ is a strongly closed bounded convex set, and the canonical inclusion $J:X\hookrightarrow X''$ has dense image, then $C$ is weakly-* closed.
Proof.
As seen above, we have $f\in X'':f(x)<\alpha<f(x_0)$ for all $x\in C$. By density of $J(X)$, we can find $\tilde f\in X$ such that $\|J(\tilde f)-f\|_{X''}$ is as small as we want. Since $C$ is bounded, $\|x\|\leq M$ for all $x\in C$. Therefore, we can make sure $|f(x)-J(\tilde f)(x)|\leq\frac{f(x_0)-\alpha}{3}$ for all $x\in C$, and also that $|f(x_0)-J(\tilde f)(x_0)|\leq\frac{f(x_0)-\alpha}{3}$. This will grant us that:
$$J(\tilde f)(x)<\frac{f(x_0)+\alpha}{2}<J(\tilde f)(x)$$
for all $x\in C$, so that $J(\tilde f)$ again separates $C$ from $\{x_0\}$, but is weak-* continuous, and thus the set $V:=[J(\tilde f)]^{-1}([\frac{f(x_0)+\alpha}{2},\infty))$ is weak-* open, disjoint from $C$, and contains $x_0$, which concludes the proof just like above.                                                                                                  $\diamond$
But $J$ only has dense image if it is surjective, since the image is closed by $J$ being an isometry, and if it is surjective, $X$ is reflexive and weak-* and weak are the same topologies. So the above is not even a generalization.
MathOverflow to the rescue
I googled "weak-*" closure vs. strong closure of convex sets and landed on this, where one suggestion is:

Suppose $X$ is not reflexive.  Then considering $X$ embedded into $X''$, we have $X \subsetneq X''$.  If $\alpha \in X'' \setminus X$, then its kernel is norm closed, weakly closed, convex, but not weak* closed.

Question
How do I prove that? I tried the following:

*

*If $K=\ker f$ is weak-*-closed, then all of its translates $K_a:=f^{-1}(\{a\})$ are, because translations are homeomorphism from weak-* to weak-*; indeed, weak-* continuity equates to the continuity of the composition with all evaluation functionals, which is obvious for translations, because, if $x\in X$, then $f\mapsto(f+f_0)(x)$ is $f(x)$ plus a constant, both weak-* continuous functions;

*This means $f^{-1}(\mathbb R\smallsetminus\{a\})$ is weak-*-open for any $a\in\mathbb R$, with $f$ our element of $X''$;

*If we could prove $f^{-1}((a,\infty))$ is weak-*-clopen in $f^{-1}(\mathbb R\smallsetminus\{a\})$, we would then have that $f$ is weak-* continuous, since open half-lines generate the topology of $\mathbb R$;

*This means that, if $\ker f$ is weak-*-closed for all $f\in X''$, the weak-* topology is the same as the weak one, meaning $X$ is reflexive, and therefore, if $X$ isn't reflexive, we can find $f\in X''\smallsetminus X$ such that $\ker f$ isn't weak-*-closed.

But how do I do step 3? Or otherwise, how do I prove $\ker f$ for $f\in X''\smallsetminus X$ is not weak-*-closed?
 A: @DavidMitra commented by posting this, which gives the following equivalent conditions of continuity:


*

*Continuity itself;

*Closed kernel;

*Non-dense kernel or $f=0$;

*Boundedness in a neighborhood of the origin.


Let's prove some implications between these statements.


*

*1=>2 is obvious;

*Closed and dense kernel implies $f=0$, so if $f\neq0$ the closed kernel must be non-dense;

*If $|f|_V|\leq M$, then pick $\frac\epsilon MV$ and you'll have $|f|\leq\epsilon$ on it, and multiplication is a homeomorphism so you have that for all $\epsilon$ there is $V_\epsilon:=\frac\epsilon MV$ such that $|f|\leq\epsilon$ on $V_\epsilon$, which means continuity at 0, but then $f(y)-f(x)=f(y-x)$ which goes to zero if $y\to x$ (aka $y-x\to0$), which means $f$ is continuous;

*The only thing left to prove is 3=>4, which I prove below following his proof; note that balanced, for a set $V$, means $\lambda V\subseteq V$ for all $\lambda:|\lambda|\leq1$.


Proof of 3=>4.


*

*$f=0$ obviously implies 4, and in fact global boundedness.

*Suppose then $\ker f$ is not dense. This means there must be
$x\notin\ker f$ and a balanced neighborhood $V$ of the origin such
that $x+V\cap\ker f=\varnothing$.

*The neightborhood can be chosen balanced by uniting any neighborhood $V$ with all the $\lambda V$ for $|\lambda|\leq1$ since, if that were impossible, we would have a sequence of points $x_n\to x$ strongly such that $f(x_n)=0$, but then $f(x)=0$ because $f$ is strongly continuous, a contradiction.

*$f(V)$ is then a balanced subset of the field $\mathbb K$ (think of
$\mathbb R$ or $\mathbb C$ here), which makes it (at least in the two
cases I mentioned) either the whole field or a bounded set.

*If it is bounded, then we have a neighborhood of the origin were $f$
is bounded, so 4 is proved.

*Otherwise, there is $y\in B_r(0):f(y)=-f(x)$, meaning $f(x+y)=0$, so
that $x+y\in B_r(x)\cap\ker f$, a set we assumed to be empty,
contradiction.


Therefore, we have the following.
Proposition
If $X$ is a Banach space and every $f\in X''$ has weak-*-closed kernel, then $X$ is reflexive.
Proof.
By the above, we have that $f\in X''$ implies $f$ is weak-*-continuous, but then the weak-* topology must contain the weak topology, which is the coarsest of those that make all $f\in X''$ continuous, yet we know the reverse inclusion holds, thus making the two topologies coincide. This means the closed unit ball of $X'$ is weakly compact, which by Kakutani's theorem implies $X'$ is reflexive, and then it's known that $X$ is reflexive iff $X'$ is.
Therefore, pick any non-reflexive space, and at least one $f\in X''\smallsetminus X$ is not weakly-* continuous, making its kernel convex but not weakly-* closed, yet it is surely strongly closed because $f$ is strongly continuous, meaning that:
Corollary
$X$ is reflexive iff strong and weak-* closure coincide for convex subsets of $X'$.
