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.

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One Response to Weighted L2 Spaces

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