A not Hilbert space

Question

Suppose H is the space of the bounded functions from the reals to the complex numbers. The inner product is given by : $(f,g) = \int_{\mathbb R} \dfrac{f(x)\bar g(x)}{1+x^2}\,dx$. How can I show that this space is not hilbert?

Answer 1

Remark: Notice that $\langle-,-\rangle$ does not define a scalar product if you don't set $H:=\mathcal{B}(\mathbb{R},\mathbb{C})/\sim$, where $\sim$ is the equivalence relation "to be equal a.e.". Indeed, to show a counterxample just take the function $$f(x)=\begin{cases} 1, \: \textit{ if } x\in\mathbb{Q}\\ 0, \: \textit{ if } x\in\mathbb{R} \end{cases},$$ then $\vert\vert f\vert\vert=0$.

If you take the sequence of functions define by $f_n(x)=\chi_{[0,1]}(x)(\sqrt{1+x^2})\min(x^{-1/4},n^{1/4})$, then it's simple to check that $(f_n)$ is a Cauchy sequence and that $f_n\to (\sqrt{1+x^2})f(x)$, where $f(x):=\chi_{[0,1]}(x)x^{-1/4}$.

Now notice that if a measurable function is equal a.e. to $f$, then it must be unbounded.