Is a dual space of a normed space Banach space Why

Yes, the dual space of a normed space is always a Banach space.

To see why, recall that the dual space of a normed space X is the space of all bounded linear functionals on X, denoted by X'. The norm on X' is given by

||f|| = sup{ |f(x)| : x in X and ||x|| <= 1}

We need to show that X' is complete, i.e., every Cauchy sequence in X' converges to a limit in X'.

Let (f_n) be a Cauchy sequence in X'. Then for any epsilon > 0, there exists N such that for all n, m >= N,

||f_n - f_m|| < epsilon/2

Since ||f_n - f_m|| = sup{ |f_n(x) - f_m(x)| : x in X and ||x|| <= 1}, we can choose some x_0 in X such that ||x_0|| = 1 and |f_n(x_0) - f_m(x_0)| >= ||f_n - f_m|| - epsilon/2 for all n, m >= N.

Thus, (f_n(x_0)) is a Cauchy sequence in the scalar field (either real or complex numbers). Since the scalar field is complete, there exists a limit, say L, of (f_n(x_0)).

Define a linear functional f on X by f(x) = lim f_n(x_0), as n approaches infinity. We claim that f is bounded and that f_n converges to f in X'.

To see that f is bounded, note that for any x in X with ||x|| <= 1,

|f(x)| = |lim f_n(x_0)| <= lim sup |f_n(x_0)| <= lim sup ||f_n|| ||x|| <= ||x||

where the last inequality follows from the fact that (f_n) is a Cauchy sequence. Thus, f is a bounded linear functional on X, and so f is in X'.

To see that f_n converges to f in X', note that for any x in X with ||x|| <= 1,

|f_n(x) - f(x)| = |f_n(x) - lim f_m(x_0)| = lim |f_n(x) - f_m(x_0)|

as n, m approach infinity. Since (f_n(x_0)) is a Cauchy sequence in the scalar field, we can make the right-hand side arbitrarily small by choosing n, m sufficiently large. Thus, f_n converges to f in X'.

Therefore, X' is complete, i.e., a Banach space