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A normed space $X$ is reflexive iff $X^{**}=\{g_x:x\in X\}$ where $g_x$ is bounded linear functional on $X^*$ defined by $g_x(f)=f(x)$ for any $f\in X^*$.

Let $X$ be a Hilbert space, would you help me to show that $X$ is reflexive.

One of the example is $L^2[a,b]$, the reason is its dual is $L^2$ and the second dual is $L^2$ again.

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Your example for $X = L^2[a,b]$ can be extended to apply to any Hilbert space; this is the Riesz representation theorem. – Christopher A. Wong Dec 11 '12 at 0:15
@ChristopherA.Wong: Would u help me to check my answer below? – beginner Dec 11 '12 at 0:27

Hint: use the Riesz representation theorem twice.

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Here's my answer, would you help me to check it.

Since $X^*$ is Hilbert space too, then for $g$ a bounded linear functional on $X^*$ there is $f_g \in X^*$ such that $g(f)=\langle f,f_g\rangle$ and $\|g\|=\|f_g\|$ (Riesz Representation Theorem). Since $f_g\in X^*$ then by Riesz Representation Theorem again there exist $x_{f_g}$ such that $f_g(x)=\langle x,x_{f_g}\rangle$ and $\|f_g\|=\|x_{f_g}\|$. Hence, $\|g\|=\|x_{f_g}\|$. So, $X^{**}=\{g_x:x\in X\}$

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You're pretty much there. You've basically defined a mapping $T:X^{\star\star}\rightarrow X$ where $g\mapsto x_{f_g}$; $T$ is one-to-one and onto by Riesz (twice), and indeed an isometry since $\|g\|=\|x_{f_g}\|$ as you've shown. Essentially $T$ is the composition $T=\psi_H\circ\psi_{H^*}$ where $\psi_H$, $\psi_{H^*}$ are the isometries provided by Riesz. – icurays1 Dec 11 '12 at 1:27

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