I am reading a book which states Hahn-Banach separation theorem as follows.
As far as I understand weak* topology is defined on the dual space X*.So is this a typo in the book.Should the weak* topology insted be weak topology?If not what is meant by 'closed in the weak* topology on X'?
Obviously there is a mistake - not every Banach space is a dual space (i.e. there exist Banach spaces that do not admit a weak-star topology, for example $L^1[0,1]$ does not have a predual).
My guess is that we should replace "weak-star" by "weak". The weak topology on $X$ is the one induced by the following description of convergence: $x_i\to x$ if-f $f(x_i)\to f(x)$ for all $f\in X^*$. If we do this replacement, then the classical Hahn-Banach theorem, as stated in Rudin for example, yields a functional $f:X\to\mathbb{C}$ that is continuous for the weak topology and satisfies $\sup_{x\in C}f(x)<\inf_{x\in K}f(x)$.
The only "non-trivial" (but immediate) thing to reach the theorem as stated is to show that $f$ is norm-continuous, i.e. $f\in X^*$: but indeed, if $x_i\to x$ in norm in $X$, then we also have that $x_i\to x$ weakly in $X$. But $f$ is weakly continuous, so $f(x_i)\to f(x)$. This shows that $f$ is norm-continuous, i.e. $f\in X^*$ as we wanted.