# Thread: Folland:Real Analysis, section 5.2 #24

1. ## Folland:Real Analysis, section 5.2 #24

suppose X is a Banach space, let X^\tilde and (X*)^\tilde be the natural images of X, X* in X**, X***, and let (X^\tilde)^0={F \in X***: F|_{X^\tilde}=0}. Then show that:

(X*)^\tilde+(X^\tilde)^0=X***

2. Please use [ math ] and [ /math ] tags (without the spaces) around your Latex code

3. Originally Posted by frankmelody
suppose X is a Banach space, let X^\tilde and (X*)^\tilde be the natural images of X, X* in X**, X***, and let (X^\tilde)^0={F \in X***: F|_{X^\tilde}=0}. Then show that:

(X*)^\tilde+(X^\tilde)^0=X***
This becomes easier if you use a better notation. Write $\displaystyle J:X\to X^{**}$ and $\displaystyle K:X^*\to X^{***}$ for the canonical embeddings. Let $\displaystyle J^*:X^{***}\to X^*$ be the adjoint of $\displaystyle J$, so that $\displaystyle (J^*\Phi)(x) = \Phi(Jx)\ \ (\Phi\in X^{***},\,x\in X)$.

For $\displaystyle \Phi\in X^{***}$, let $\displaystyle \Psi = K(J^*\Phi)$. Then it's just a matter of definition-chasing to check that $\displaystyle \Phi - \Psi \in (JX)^0$. That shows that $\displaystyle X^{***} = KX^* + (JX)^0$. (In fact, the sum is a direct sum.)