Let $H= \mathcal{B}(\mathbb{R},\mathbb{C})$ be the space of measurable, bounded, complex valued functions on $\mathbb{R};$ define
$$(f,g) = \int_{\mathbb R} \dfrac{f(x)\bar g(x)}{1+x^2}\,dx$$
I would like to see that this is not a Hilbert space. I found this problem here, but it is now closed and I don't understand the solution given..
Can you help me understand that solution or give a proof?
As mentioned in the earlier post this is not an inner product unless you identify two functions when they are equal almost everywhere. Let us assume that we have done that.
Let $f_n(x)=x^{1/3}$ if $0<x <n$ and $0$ for all other $x$. Let us show that this sequence is Cauchy. We have $\int \frac {|f_n(x)-f_m(x)|^{2}} {1+x^{2}}dx=\int_n^{m} \frac {x^{2/3}} {1+x^{2}}$ if $n <m$. This last integral does not exceed $\int_n^{m} \frac 1 {x^{4/3}}dx=3(\frac 1{n^{1/3}}- \frac 1{m^{1/3}}) \to 0$. So we have proved that $(f_n)$ is Cauchy. Suppose this sequence converges to some $f$. Then $\int {|f_n-f|^{2}} d\mu(x) \to 0$ where $\mu$ is defined by $\mu (E)=\int_E \frac 1{1+x^{2}}dx$. By a basic result in measure theory this implies that there is a subsequence $(f_{n_k})$ which converges almost everywhere (w.r.t $\mu$, hence also w.r.t Lebesgue measure). Now let $x>0$. Then $0<x <n_k$ for $k$ sufficiently large so we must have $f(x)=\lim f_{n_k}(x)= x^{1/3}$. But then $f$ is not bounded so it does not belong to the space under consideration.