# The dual $H'$ of a Hilbert Space $H$ is a Hilbert Space

Let H be a Hilbert Space. Show that the dual space $H'$ of $H$ is a Hilbert Space with inner product $\langle \cdot, \cdot \rangle_1$ defined by $$\langle f_z , f_v \rangle_1 = \overline{ \langle z,v\rangle}=\langle v,z\rangle,$$ where $f_z(x)=\langle x,z\rangle$, with $\langle \cdot, \cdot\rangle$ is the inner product in $H$.

I have already shown that $\langle\cdot , \cdot\rangle_1$ is inner product. Now I need to prove that $H'$ is complete, so I started this way: we have that $H'$ is an inner product space and the metric $d:H'\times H'\rightarrow \mathbb{R}$ defined by $$d(f_z, f_v)= \sqrt{\langle f_z -f_v,f_z -f_v\rangle_1}.$$ Let $({f_z}_n)_{n\in\mathbb{N}}$ be a arbitrary Cauchy sequence in $H'$, that is $$\forall \epsilon >0, \: \exists \: n_0 \in \mathbb{N} \:;\: m,n>n_0 \: \Rightarrow \: d({f_z}_n,{f_z}_m)<\epsilon.$$

I can write that ${f_z}_n=\langle x,z_n\rangle$? Where $(z_n)_{n\in\mathbb{N}}$ is a sequence in H.

How to continue to prove that such a sequence converges?

• I advice you to use \langle $\langle$ and \rangle $\rangle$ instead of < and > to denote the inner product also see here Dec 7 '17 at 18:46

The map $\phi : H \to H'$ given by $\phi(v) = f_v$, where $f_v(x) = \langle x, v\rangle$, for $x \in H$ is an antilinear bijective isometry. Bijectivity and norm-preservation both follow from the Riesz Representation Theorem, and antilinearity can be easily verified:

$$\phi(\alpha v + \beta w)(x) = f_{\alpha v + \beta w}(x) = \langle x, \alpha v + \beta w \rangle = \overline{\alpha}\langle x, v\rangle + \overline{\beta}\langle x, w\rangle = \overline{\alpha}f_v(x) + \overline{\beta}f_w(x) = \big(\overline{\alpha}\phi(v) + \overline{\beta}\phi(w)\big)(x)$$

Hence, $\phi(\alpha v + \beta w) = \overline{\alpha}\phi(v) + \overline{\beta}\phi(w)$.

Now, as you can show, if two spaces are "antilinearly isometric", then one is complete if and only if the other one is complete.

Indeed, let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $H$. We have $f_n = \phi(x_n)$ for some $x_n \in H$.

The sequence $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $H$:

$$\|x_m - x_n\| = \|\phi(x_m - x_n)\| = \|\phi(x_m) - \phi(x_n)\| = \|f_m - f_n\| \xrightarrow{m, n \to\infty} 0$$

Since $H$ is complete, $(x_n)_{n=1}^\infty$ converges. Set $x_n \xrightarrow{n\to\infty} x \in H$.

We claim that $f_n \xrightarrow{n\to\infty} \phi(x)$ in $H'$. Indeed:

$$\|\phi(x) - f_n\| = \|\phi(x) - \phi(x_n)\| = \|\phi(x - x_n)\| = \|x - x_n\| \xrightarrow{n\to\infty} 0$$

Hence, $H'$ is complete.

So $f_{z_{n}}-f_{z_{m}}=f_{z_{n}-z_{m}}$ and hence if $d(f_{z_{n}},f_{z_{m}})<\epsilon$, this means $\left<f_{z_{n}-z_{m}},f_{z_{n}-z_{m}}\right>\leq\epsilon^{2}$, or $\left<z_{n}-z_{m},z_{n}-z_{m}\right><\epsilon^{2}$, the latter is $\|z_{n}-z_{m}\|<\epsilon$, as $H$ is complete, there is some $z\in H$ such that $z_{n}\rightarrow z$. Now it is routine to prove that $f_{z_{n}}\rightarrow f_{z}$.

Note that $d(f_z,f_v)=\|v-z\|^2$, where $\|\cdot\|$ is the norm on $H$ induced by $\langle\cdot,\cdot\rangle$. Use this to show that if $\{f_{z_n}\}$ is Cauchy then $\{z_n\}$ is Cauchy. Then prove that if $z_n\to z$ then $f_{z_n}\to f_z$.

For any normed vs $$X$$ the dual $$X^*$$ is complete. In this case, the isometric mapping ensures that the norm induced by the inner product is the norm of the dual space.

So you could show it more generally and then arrive at your claim.