# 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 – Frederik vom Ende 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 $f_{z_n}\to f_z$.