## Weighted L2 Spaces

Let $\mathcal{H}, \mathcal{H}'$ be Hilbert spaces with inner products $(\cdot,\cdot)_{\mathcal{H}}$ and $(\cdot,\cdot)_{\mathcal{H}'}$, respectively, and corresponding norms $\left\|\cdot\right\|_{\mathcal{H}}$ and $\left\|\cdot\right\|_{\mathcal{H}'}$. A linear transformation $U: \mathcal{H}\rightarrow\mathcal{H}'$ is called unitary if

1. $U$ is bijective.
2. $\left\|Uf\right\|_{\mathcal{H}'}=\left\|f\right\|_{\mathcal{H}}$ for all $f \in \mathcal{H}$.

$U$ is bijective, and the second property shows that $\left\|U^{-1}f\right\|_{\mathcal{H}} = \left\|U(U^{-1}f)\right\|_{\mathcal{H}'} = \left\|f\right\|_{\mathcal{H}}$, hence $U^{-1}: \mathcal{H}' \rightarrow \mathcal{H}$ is unitary.

Using the polarization identity for inner product spaces, we obtain the following equivalence for the second property.

Lemma 1. $U: \mathcal{H} \rightarrow \mathcal{H}'$ is a unitary mapping if and only if  $U$ is a bijective linear transformation satisfying

$\displaystyle (Uf,Ug)_{\mathcal{H}'} = (f,g)_{\mathcal{H}'}, \indent \forall f,g \in \mathcal{H}$

We say that two Hilbert spaces are unitarily equivalent or unitarily isomorphic if there exists a unitary mapping $U: \mathcal{H}\rightarrow \mathcal{H}'$.

We now turn to weighted $L^{2}$ spaces, which will talk more about when we deal with Sturm-Liouville problems. Let $\eta: [a,b] \rightarrow \mathbb{R}$ be a strictly positive continuous function. Define the $\mathcal{H}_{\eta} = L^{2}([a,b],\eta)$ to be the space of all measurable functions $f: [a,b] \rightarrow \mathbb{C}$ such that

$\displaystyle \int_{a}^{b}\left|f(t)\right|^{2}\eta(t)dt < \infty$

I claim that we can define an inner product on $\mathcal{H}_{n}$ by

$\displaystyle (f,g)_{\eta} = \int_{a}^{b}f(t)\overline{g(t)}\eta(t)dt$

The linearity in the first argument and the conjugate linearity in the second argument follows from the observation that $(f,g)_{\eta} = (f,g\eta)$, where $(\cdot,\cdot)$ denotes the standard $L^{2}$ inner product. That $(f,f)_{\eta} \geq 0$ follows from the nonnegativity of $\left|f\right|^{2}\eta$. Equality holds in the preceding inequality if and only if $f = 0$ since $\eta(t) \geq \epsilon > 0$ for all $t \in [a,b]$ by Weierstrass’ extreme value theorem.

To see that $\mathcal{H}_{\eta}$ is complete and therefore a Hilbert space, let $(f_{n})_{n=1}^{\infty}$ be a Cauchy sequence with respect to the norm $\left\|\cdot\right\|_{\eta}$. It suffices to show that $(f_{n})$ has a subsequence which converges to a function $f \in L^{2}([a,b],\eta)$. We can choose a strictly increasing sequence of indices $n_{k} < n_{k+1}$ such that

$\displaystyle \left\|f_{n_{k}}-f_{n_{k+1}}\right\|_{\eta} \leq \frac{1}{2^{k}}$

Note that by Minkowski’s inequality,

$\displaystyle \int_{a}^{b}\left(\left|f_{n_{1}}\right|\eta^{\frac{1}{2}}+\sum_{k=1}^{\infty}\left|f_{n_{k+1}}-f_{n_{k}}\right|\eta^{\frac{1}{2}}\right)^{2}dt \leq \left(\left\|f_{n_{1}}\eta^{\frac{1}{2}}\right\|_{L^{2}}+\sum_{k=1}^{\infty}\left\|(f_{n_{k+1}}-f_{n_{k}})\eta^{\frac{1}{2}}\right\|_{L^{2}}\right)^{2}$

By the monotone converge theorem, $\left|f_{n_{1}}\eta^{\frac{1}{2}}\right|+\sum_{k=1}^{\infty}\left|(f_{n_{k+1}}-f_{n_{k}})\eta^{\frac{1}{2}}\right|$ converges a.e. and is in $L^{2}([a,b])$. By the dominated convergence theorem,

$\displaystyle f_{n_{k+1}}\eta^{\frac{1}{2}}=f_{n_{1}}\eta^{\frac{1}{2}}+\sum_{k=1}^{N}(f_{n_{k+1}}-f_{n_{k}})\eta^{\frac{1}{2}} \longrightarrow g\in L^{2}([a,b])$

Set $f := g\eta^{-\frac{1}{2}}$. It is clear that $f\in \mathcal{H}_{\eta}$ and

$\displaystyle 0 = \lim_{k\rightarrow\infty}\int_{a}^{b}\left|f_{n_{k+1}}\eta^{\frac{1}{2}}-g\right|^{2}dt=\lim_{k\rightarrow\infty}\int_{a}^{b}\left|f_{n_{k+1}}-f\right|^{2}\eta dt=\lim_{k\rightarrow\infty}\left\|f_{n_{k+1}}-f\right\|_{\eta}^{2}$

We summarize our results in the following theorem.

Theorem 2. For a fixed strictly positive, continuous function $\eta: [a,b] \rightarrow \mathbb{R}$, $L^{2}([a,b],\eta)$ is a Hilbert space.

Above, we restricted ourselves to the case where $\eta$ was strictly positive and continuous, but we can relax this hypothesis to $\eta$ is a strictly positive and measurable on $[a,b]$. The only nontrivial verification is that $(f,f)_{\eta} = 0$ implies that $f = 0$ a.e. Recall that for a nonnegative measurable function $g$,

$\displaystyle \int_{a}^{b}g(t)dt = 0 \Longleftrightarrow g = 0$ a.e.

Applying this result to $g=\left|f\right|^{2}\eta$, we see that

$\displaystyle (f,f)_{\eta} = 0 \Longleftrightarrow \left|f\right|^{2}\eta = 0$ a.e. $\displaystyle\Longleftrightarrow f = 0$ a.e.

where the penultimate equivalence follows from the strict positivity of $\eta$.

It might not surprise the reader that $L^{2}([a,b],\eta)$ and $L^{2}([a,b])$ are unitarily equvalent. Define a linear transformation

$\displaystyle U: L^{2}([a,b]) \rightarrow L^{2}([a,b],\eta), f \mapsto f\eta^{-\frac{1}{2}}$

It is clear that $U$ is a linear transformation and

$\displaystyle \left\|Uf\right\|_{\eta}^{2} = \int_{a}^{b}\left|f\eta^{-\frac{1}{2}}\right|^{2}\eta dt = \int_{a}^{b}\left|f\right|^{2}\eta^{-1}\eta dt = \left\|f\right\|_{L^{2}}^{2},$

from which injectivity also follows. To see that $U$ is surjective, for $g \in L^{2}([a,b],\eta)$, define $f := g\eta^{\frac{1}{2}}$. We just need to verify that $f\in L^{2}([a,b],\eta)$. Indeed,

$\displaystyle \left\|f\right\|_{L^{2}}^{2} = \int_{a}^{b}\left|g\eta^{\frac{1}{2}}\right|^{2}dt = \int_{a}^{b}\left|g\right|^{2}\eta dt = \left\|g\right\|_{\eta}^{2} < \infty$

We summarize this result with our last theorem.

Theorem 3. For a fixed strictly positive, measurable function $\eta$, $L^{2}([a,b])$ and $L^{2}([a,b],\eta)$ are unitarily equivalent.