# Modified $L^2$ space is a Hilbert space

I'm trying to show that the space ($X$) of measurable functions on $[a,b]$ such that for a fixed, positive, function $h$, and $\forall f \in X$, $\int_a^b |f(x)|^2 h(x) < \infty$ is a Hilbert space with respect to the inner product $\langle f,g \rangle= \int_a^b f(x) {g^*(x)} h(x)dx$.

I'm mainly wondering about showing that this space is complete and separable. I have the corresponding proofs for $L^2([a,b])$ which I could just modify, but I am wondering if there is an easier way to prove this since the space is so similar to $L^2[a,b]$

Alternatively, i.e without considering $L^2$ spaces with respect to measures different from the Lebesgue measure, we can notice that $$X = \left\{f : f. h^{\frac{1}{2}} \in L^{2}[a, b] \right\}$$ and $$\forall f \in X, ||f||_{X} = ||f. h^{\frac{1}{2}}||_{L^{2}}$$ From there, it is not hard to prove that $X$ is both complete and separable.

• For completeness:
Let $(f_{n})_{n \geq 1}$ be a Cauchy sequence in $X$.
Then $\left(f_{n}. h^{\frac{1}{2}}\right)_{n \geq 1}$ is a Cauchy sequence in $L^{2}[a, b]$ and hence, since $L^{2}[a, b]$ is complete, it converges to some $g \in L^{2}[a, b]$ with respect to $||.||_{L^{2}}$.
Therefore, $\frac{g}{h^{\frac{1}{2}}} \in X$ and $(f_{n})_{n \geq 1}$ converges to $\frac{g}{h^{\frac{1}{2}}}$ with respect to $||.||_{X}$.

• $X$ is separable:
Since $L^{2}[a, b]$ is separable, there is a dense sequence $(f_{n})_{n \geq 1}$ in $L^{2}[a, b]$ with respect to $||.||_{L^{2}}$.
Then $\left(\frac{f_{n}}{h^{\frac{1}{2}}}\right)_{n \geq 1}$ is a dense sequence in $X$ with respect to $||.||_{X}$.

• This is not totally true. You just have $X = \{f \mid f \, h^{1/2} \in L^2\}$. $f$ itself may not belong to $L^2$. – gerw May 7 '18 at 15:14
• Thank you for spotting the mistake. I've edited my post. – v_lentin May 7 '18 at 15:19

If you set $\mu (A) =\int_A h(x) dx$ for the all Lebesgue measurable sets $A\subset [a,b],$ then $\mu$ is a positive measure and $$\int_a^b f(x)g^* (x) h(x) dx =\int_{[a,b]} f(t)g(t) \mu (dt)$$ hence the space that you consider becomes a space $L^2 (\mu )$ but all those spaces are complete. Proof of this fact you can find in almost any book of functional analysis.

• What does $L^2(\mu)$ mean? sorry this is a new topic for me – dimebucker May 7 '18 at 13:00