Kernel of functional on dual space is weak-* closed iff it is evaluation functional 
Let $X$ be a Banach space and $f \in X^{**}$. Show that $\ker f$ is weak-* closed if and only if $f = i_x$ for some $x \in X$, where $i_x\in X^{**}$ is the evaluation functional $i_x(g)\equiv g(x)$ ($g\in X^*$).

I’ve seen this topic, but in this problem the author considers the case with a set of kernels. Can I reduce my problem to this one?
 A: Suppose that $\phi\in X^{**}$ and that $\phi=i_x$ for some $x\in X$ (I prefer to reserve the notation $f$ for elements of $X^*$). Since the weak-star topology on $X^*$ is, by definition, the weakest topology that renders all $\{i_y\mid y\in X\}$ continuous, it follows, in particular, that $\phi$ is weak-star continuous and its kernel is weak-star closed.
Conversely, suppose that $\ker\phi$ is weak-star closed. If $\phi$ identically vanishes, then $\phi=i_0$, so assume that there exists some $f\in X^*$ such that $\phi(f)\neq 0$. Therefore, $f\in X^*\setminus\ker\phi$. Now, since $X^*\setminus\ker\phi$ is open in the weak-star topology, the way the weak-star topology is generated guarantees the existence of some $\varepsilon>0$, $n\in\mathbb N$, and $x_1,\ldots,x_n\in X$ such that $$f\in\bigcap_{j=1}^n\left\{g\in X^*\mid|i_{x_j}(g)-i_{x_j}(f)|<\varepsilon\right\}\subseteq X^*\setminus\ker\phi.$$
Let the intersection appearing in the formula above be denoted as $U$. I now claim that $$\bigcap_{j=1}^n\ker i_{x_j}\subseteq\ker\phi.\tag{$\clubsuit$}$$ Indeed, if $g\in X^*$ is such that $g(x_1)=\cdots=g(x_n)=0$, then $f+\mu g\in U\subseteq X^*\setminus\ker\phi$ for all $\mu\in\mathbb F$ (where $\mathbb F\in\{\mathbb R,\mathbb C\}$ is the underlying field). This implies that $\phi(f+\mu g)=\phi(f)+\mu\phi(g)\neq 0$ for any $\mu\in\mathbb F$, which is only possible if $\phi(g)=0$; otherwise, one could take $\mu=-\phi(f)/\phi(g)$. Therefore, $g\in\ker\phi$, as claimed.
Finally, $(\clubsuit)$ implies that $\phi\in\operatorname{span}\{i_{x_1},\ldots,i_{x_n}\}$. To see this, define $\Xi:X^*\to\mathbb F^n$ as $$\Xi(g)\equiv(i_{x_1}(g),\ldots,i_{x_n}(g))\quad\text{for each $g\in X^*$.}$$ Note that if some $g,h\in X^*$ satisfy $$(i_{x_1}(g),\ldots,i_{x_n}(g))=(i_{x_1}(h),\ldots,i_{x_n}(h)),$$ then $i_{x_j}(g-h)=0$ for all $j\in\{1,\ldots,n\}$, so that $(\clubsuit)$ implies that $\phi(g)=\phi(h)$. Therefore, the function $\Psi:\Xi(X^*)\to\mathbb F$, defined as $$\Psi(i_{x_1}(g),\ldots,i_{x_n}(g))\equiv\phi(g)\quad\text{for $g\in X^*$},$$ is well-defined. It is easy to check that it is also a linear function mapping from $\Xi(X^*)$ (which is a linear subspace of $\mathbb F^n$) to $\mathbb F$, so that it can be extended to a linear function mapping from $\mathbb F^n$ to $\mathbb F$ (details provided below). This implies that there exist $\lambda_1,\ldots,\lambda_n\in\mathbb F$ such that, for each $g\in X^*$, $$\phi(g)=\sum_{j=1}^n\lambda_j i_{x_j}(g)=\sum_{j=1}^n\lambda_ jg(x_j)=g\left(\sum_{j=1}^n\lambda_ jx_j\right).$$ Hence, $\phi=i_{\lambda_1x_1+\cdots+\lambda_nx_n}$, which completes the proof.

Let me explain how to extend $\Psi:\Xi(X^*)\to\mathbb F$ to a linear function mapping from $\mathbb F^n$ to $\mathbb F$ and where the coefficients $\lambda_1,\ldots,\lambda_n$ come from. Let $\{e_1,\ldots,e_m\}\subseteq\Xi(X^*)$ be a basis for the subspace $\Xi(X^*)$, where $m\in\{1,\ldots,n\}$. (The case in which $\Xi(X^*)=\{(0,\ldots,0)\}$ is pretty trivial.) It is possible to extend it to a basis $\{e_1,\ldots,e_m,e_{m+1},\ldots,e_n\}\subseteq\mathbb F^n$ of $\mathbb F^n$. Each $z\in\mathbb F^n$ has a unique representation $$z=\sum_{j=1}^n\alpha_j^ze_j\quad\text{for some $\alpha_1^z,\ldots,\alpha_n^z\in\mathbb F$}.$$ For each $z\in\mathbb F^n$, define $$\Theta(z)\equiv\sum_{j=1}^m\alpha_j^z\Psi(e_j).$$ It is not difficult to check that $\Theta:\mathbb F^n\to\mathbb F$ is linear and satisfies $\Theta(z)=\Psi(z)$ for $z\in\Xi(X^*)$. The coefficients $\lambda_1,\ldots,\lambda_n$ can now be defined as
\begin{align*}
\lambda_1\equiv&\;\Theta(1,0,\ldots,0,0),\\
\lambda_2\equiv&\;\Theta(0,1,\ldots,0,0),\\
\vdots&\;\\
\lambda_{n-1}\equiv&\;\Theta(0,0,\ldots,1,0),\\
\lambda_n\equiv&\;\Theta(0,0,\ldots,0,1).
\end{align*}
This way, one has, for each $g\in X^*$,
\begin{align*}
\phi(g)=&\;\Psi(i_{x_1}(g),i_{x_2}(g),\ldots,i_{x_{n-1}}(g),i_{x_n}(g))=\Theta(i_{x_{1}}(g),i_{x_2}(g),\ldots,i_{x_{n-1}}(g),i_{x_n}(g))\\
=&\;\Theta\big(i_{x_{1\phantom{-n}}}(g)\times(1,0,\ldots,0,0)+i_{x_2}(g)\times(0,1,\ldots,0,0)+\cdots+\\
&\;\phantom{\Theta\big(}i_{x_{n-1}}(g)\times(0,0,\ldots,1,0)+i_{x_n}(g)\times(0,0,\ldots,0,1)\big)\\
=&\;i_{x_{1\phantom{-n}}}(g)\times\Theta(1,0,\ldots,0,0)+i_{x_2}(g)\times\Theta(0,1,\ldots,0,0)+\cdots+\\
&\;i_{x_{n-1}}(g)\times\Theta(0,0,\ldots,1,0)+i_{x_{n}}(g)\times\Theta(0,0,\ldots,0,1)\\
=&\;\sum_{j=1}^n i_{x_j}(g)\times\lambda_j.
\end{align*}
