Show that a space $H$ is Hilbert. Let
\begin{equation}
H = \{f \in L^2(0,\infty):\int_0^\infty f^2(x)e^{-x}dx \lt\infty\}
\end{equation}

Show that $H$ is a Hilbert space with scalar product
\begin{equation}
\langle f,g \rangle = \int_0^\infty f(x)g(x)e^{-x}dx
\end{equation}

I was trying to show that it is complete about the norm coming from scalar product, but I think it's way too difficult. Instead, I was thinking to define a Unitary isomorphism between $H$ and $L^2(0,\infty)$ like this:
\begin{equation}
U:H\rightarrow L^2(0,\infty)
\end{equation}

such that $\langle U(f),U(g)\rangle_{L^2}$ = $\langle f,g\rangle_{H}$.

The idea is that the scala product on $H$ is really similar to $L^2$ 's one.
Any suggestions on how to define it properly?
I think this is false. $\{I_{(0,n)}\}$ is a Cauchy sequence in $H$ which is not convergent. If this sequence converges in $H$ then some sequence converges almost everywhere w.r.t. $e^{-x}dx$ hence w.r.t. Lebesgue measure and the limit has to be $1$ almost everywhere w.r.t. Lebesgue measure. But then the limit is not in $L^{2}(0,\infty)$, hence not in $H$.
Ok this is what I got in the end:

Again, we have

\begin{equation}
H = \{f \in L^2(0,\infty): \int_0^\infty f^2(x)e^{-x}dx &lt; \infty\}
\end{equation}
and
\begin{equation}
\langle f,g \rangle_{H} = \int_0^\infty f(x)g(x)e^{-x}dx
\end{equation}

So, (I guess),
\begin{equation}
||f||_H = \sqrt{\langle f,f \rangle_{H}} = \left(\int_0^\infty f^2(x)e^{-x}dx \right)^{1/2}
\end{equation}

Taking $f_n = \chi_{(0,n)}$ we have $f_n \in L^2(0,\infty) \forall n$ and it is Cauchy in $H$ because, assuming $n &gt;m$ we could consider
\begin{equation}
||\chi_{(0,n)}-\chi_{(0,m)}||_H = ||\chi_{(m,n)}||_H = \int_m^n e^{-x}dx = -e^{-n}+e^{-m}&lt;\epsilon
\end{equation}
For $m,n$ sufficiently large. And also
\begin{equation}
||\chi_{(0,n)}-\chi_{(0,\infty)}||_H = \int_0^\infty (\chi_{(0,n)}-\chi_{(0,\infty)})^2e^{-x}dx &lt; \infty
\end{equation}

Which step could be wrong?

I'm sorry for my mistakes, I'm here to learn, and also to find people to talk over with.

Thank you