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Section 9.6 The Hille-Yosida and Stone Theorems for self-adjoint operators

Remark 9.6.1.

Let \(A\) be self-adjoint with spectral measure \(E\text{,}\) and let

\begin{equation*} T(t)=e^{tA}=\int_{(-\infty,0]} e^{t\lambda}\,dE(\lambda). \end{equation*}

Then

\begin{equation*} T(t)-I=\int_{(-\infty,0]} (e^{t\lambda}-1)\,dE(\lambda) \end{equation*}

by the functional calculus.

Now fix \(x\in H\text{,}\) and define the scalar measure

\begin{equation*} \mu_x(B):=\langle E(B)x,x\rangle \end{equation*}

for Borel sets \(B\subset \mathbb R\text{.}\) This is a finite positive measure, since

\begin{equation*} \mu_x(\mathbb R)=\langle E(\mathbb R)x,x\rangle=\langle Ix,x\rangle=\|x\|^2. \end{equation*}

Set

\begin{equation*} f_t(\lambda):=e^{t\lambda}-1. \end{equation*}

Then

\begin{equation*} (T(t)-I)x=f_t(A)x. \end{equation*}

Therefore

\begin{equation*} \|T(t)x-x\|^2=\|f_t(A)x\|^2. \end{equation*}

To compute this norm, use the basic Hilbert-space functional calculus identity

\begin{equation*} \langle f(A)x,g(A)x\rangle = \int_{\mathbb R} f(\lambda)\overline{g(\lambda)}\,d\mu_x(\lambda) \end{equation*}

for bounded Borel functions \(f,g\text{.}\) Taking \(g=f_t\text{,}\) we get

\begin{equation*} \|f_t(A)x\|^2 = \langle f_t(A)x,f_t(A)x\rangle = \int_{\mathbb R} |f_t(\lambda)|^2\,d\mu_x(\lambda). \end{equation*}

Hence

\begin{equation*} \|T(t)x-x\|^2 = \int_{\mathbb R} |e^{t\lambda}-1|^2\,d\mu_x(\lambda). \end{equation*}

Since the spectrum of \(A\) is contained in \((-\infty,0]\text{,}\) one may equally write this as

\begin{equation*} \|T(t)x-x\|^2 = \int_{(-\infty,0]} |e^{t\lambda}-1|^2\,d\mu_x(\lambda). \end{equation*}

To justify the identity used above, let \(f\) be a bounded Borel function. Then

\begin{equation*} f(A)^*=\overline{f}(A). \end{equation*}

Hence

\begin{equation*} \|f(A)x\|^2 = \langle f(A)x,f(A)x\rangle = \langle f(A)^*f(A)x,x\rangle. \end{equation*}

By the multiplicative property of the functional calculus,

\begin{equation*} f(A)^*f(A)=\overline{f}(A)f(A)=(|f|^2)(A). \end{equation*}

Therefore

\begin{equation*} \|f(A)x\|^2 = \langle (|f|^2)(A)x,x\rangle. \end{equation*}

By definition of the scalar spectral measure \(\mu_x\text{,}\)

\begin{equation*} \langle h(A)x,x\rangle=\int_{\mathbb R} h(\lambda)\,d\mu_x(\lambda) \end{equation*}

for every bounded Borel function \(h\text{.}\) Applying this with \(h=|f|^2\) gives

\begin{equation*} \|f(A)x\|^2=\int_{\mathbb R}|f(\lambda)|^2\,d\mu_x(\lambda). \end{equation*}

Assume first that \(A\) is self-adjoint and \(A\le 0\text{.}\) By the spectral theorem, there is a projection-valued measure \(E\) on \((-\infty,0]\) such that

\begin{equation*} A=\int_{(-\infty,0]} \mu \, dE(\mu). \end{equation*}

For \(t\ge 0\text{,}\) define

\begin{equation*} T(t)=e^{tA}:=\int_{(-\infty,0]} e^{t\mu}\, dE(\mu). \end{equation*}

Since \(\mu\le 0\text{,}\) we have \(0\le e^{t\mu}\le 1\text{,}\) so

\begin{equation*} \|T(t)\| \le \sup_{\mu\le 0} e^{t\mu}\le 1. \end{equation*}

Thus each \(T(t)\) is a contraction.

The functional calculus gives

\begin{equation*} T(t)T(s)=\int e^{t\mu}\,dE(\mu)\int e^{s\mu}\,dE(\mu) =\int e^{(t+s)\mu}\,dE(\mu)=T(t+s), \end{equation*}

and clearly \(T(0)=I\text{.}\) Hence \(\{T(t)\}_{t\ge 0}\) is a semigroup.

We next show strong continuity. For \(x\in H\text{,}\) let

\begin{equation*} \mu_x(B)=\langle E(B)x,x\rangle. \end{equation*}

Then

\begin{equation*} \|T(t)x-x\|^2=\int_{(-\infty,0]} |e^{t\mu}-1|^2\,d\mu_x(\mu). \end{equation*}

As \(t\downarrow 0\text{,}\) we have \(e^{t\mu}\to 1\) pointwise, and \(|e^{t\mu}-1|^2\le 4\text{.}\) Since

\begin{equation*} \mu_x(\mathbb{R})=\|x\|^2<\infty, \end{equation*}

dominated convergence implies

\begin{equation*} \|T(t)x-x\|\to 0. \end{equation*}

Therefore \(\{T(t)\}_{t\ge 0}\) is strongly continuous.

We now identify the generator. Let \(G\) denote the generator of the semigroup. If \(x\in D(A)\text{,}\) then

\begin{equation*} \frac{T(t)x-x}{t} =\int \frac{e^{t\mu}-1}{t}\,dE(\mu)x. \end{equation*}

Also,

\begin{equation*} Ax=\int \mu\,dE(\mu)x. \end{equation*}

Therefore

\begin{equation*} \left\|\frac{T(t)x-x}{t}-Ax\right\|^2 =\int \left|\frac{e^{t\mu}-1}{t}-\mu\right|^2\,d\mu_x(\mu). \end{equation*}

Pointwise, \(\dfrac{e^{t\mu}-1}{t}\to \mu\) as \(t\downarrow 0\text{.}\) Also, for \(\mu\le 0\text{,}\) the mean value theorem gives

\begin{equation*} \left|\frac{e^{t\mu}-1}{t}\right|\le |\mu|. \end{equation*}

Hence

\begin{equation*} \left|\frac{e^{t\mu}-1}{t}-\mu\right|\le 2|\mu|, \end{equation*}

so the integrand is dominated by \(4\mu^2\text{,}\) which is integrable because \(x\in D(A)\text{.}\) By dominated convergence,

\begin{equation*} \frac{T(t)x-x}{t}\to Ax. \end{equation*}

Thus \(x\in D(G)\) and \(Gx=Ax\text{,}\) so \(A\subset G\text{.}\)

To compute the resolvent, fix \(\lambda>0\) and define

\begin{equation*} R_\lambda=\int_0^\infty e^{-\lambda t}T(t)\,dt. \end{equation*}

Since \(\|T(t)\|\le 1\text{,}\) this integral converges in operator norm and

\begin{equation*} \|R_\lambda\|\le \int_0^\infty e^{-\lambda t}\,dt=\frac{1}{\lambda}. \end{equation*}

Using the spectral theorem and Fubini's theorem,

\begin{equation*} R_\lambda =\int_0^\infty e^{-\lambda t}\left(\int e^{t\mu}\,dE(\mu)\right)\,dt =\int \left(\int_0^\infty e^{-t(\lambda-\mu)}\,dt\right)dE(\mu). \end{equation*}

Since \(\mu\le 0\) and \(\lambda>0\text{,}\) we have \(\lambda-\mu>0\text{,}\) so

\begin{equation*} \int_0^\infty e^{-t(\lambda-\mu)}\,dt=\frac{1}{\lambda-\mu}. \end{equation*}

Therefore

\begin{equation*} R_\lambda=\int \frac{1}{\lambda-\mu}\,dE(\mu)=(\lambda I-A)^{-1}. \end{equation*}

It follows that

\begin{equation*} \|(\lambda I-A)^{-1}\| \le \sup_{\mu\le 0}\frac{1}{\lambda-\mu} =\frac{1}{\lambda}. \end{equation*}

This is the Hille--Yosida resolvent estimate in the self-adjoint case.

Since the generator \(G\) has resolvent \((\lambda I-A)^{-1}\) for every \(\lambda>0\text{,}\) we conclude that \(G=A\text{.}\) Thus \(A\) generates the strongly continuous contraction semigroup \(e^{tA}\text{.}\)

Conversely, assume that \(A\) is self-adjoint and satisfies

\begin{equation*} \|(\lambda I-A)^{-1}\| \le \frac{1}{\lambda}\qquad (\lambda>0). \end{equation*}

Because \(A\) is self-adjoint,

\begin{equation*} \|(\lambda I-A)^{-1}\| =\sup_{\mu\in \sigma(A)} \frac{1}{|\lambda-\mu|}. \end{equation*}

If there were some \(\mu_0>0\) in \(\sigma(A)\text{,}\) then taking \(\lambda\) close to \(\mu_0\) from the resolvent set side would force \(\|(\lambda I-A)^{-1}\|\) to be arbitrarily large, contradicting the estimate \(\|(\lambda I-A)^{-1}\|\le 1/\lambda\text{.}\) Hence

\begin{equation*} \sigma(A)\subset (-\infty,0], \end{equation*}

so \(A\le 0\text{.}\)

Applying the first part of the proof, \(A\) generates the contraction semigroup \(T(t)=e^{tA}\text{.}\) This proves the equivalence.

Since \(A\) is self-adjoint, the spectral theorem gives a projection-valued measure \(E\) on \(\mathbb R\) such that

\begin{equation*} A=\int_{\mathbb R} \lambda\,dE(\lambda). \end{equation*}

For each \(t\in\mathbb R\text{,}\) define

\begin{equation*} U(t)=\int_{\mathbb R} e^{it\lambda}\,dE(\lambda). \end{equation*}

Because \(|e^{it\lambda}|=1\) for every \(\lambda\in\mathbb R\text{,}\) the functional calculus implies that \(U(t)\) is bounded and

\begin{equation*} \|U(t)\|\le \sup_{\lambda\in\mathbb R}|e^{it\lambda}|=1. \end{equation*}

Also,

\begin{equation*} U(t)^*=\int_{\mathbb R} \overline{e^{it\lambda}}\,dE(\lambda) =\int_{\mathbb R} e^{-it\lambda}\,dE(\lambda)=U(-t). \end{equation*}

Therefore

\begin{equation*} U(t)^*U(t)=U(-t)U(t) =\int_{\mathbb R} e^{-it\lambda}e^{it\lambda}\,dE(\lambda) =\int_{\mathbb R}1\,dE(\lambda)=I. \end{equation*}

Similarly, \(U(t)U(t)^*=I\text{,}\) so \(U(t)\) is unitary.

The multiplicative property of the functional calculus gives

\begin{equation*} U(t)U(s) =\int_{\mathbb R} e^{it\lambda}\,dE(\lambda)\int_{\mathbb R} e^{is\lambda}\,dE(\lambda) =\int_{\mathbb R} e^{i(t+s)\lambda}\,dE(\lambda)=U(t+s). \end{equation*}

Also \(U(0)=I\text{.}\) Hence \(\{U(t)\}_{t\in\mathbb R}\) is a unitary group.

It remains to prove strong continuity. Fix \(x\in H\text{,}\) and define the finite measure

\begin{equation*} \mu_x(B)=\langle E(B)x,x\rangle. \end{equation*}

Then

\begin{equation*} \|U(t)x-U(s)x\|^2 =\int_{\mathbb R}|e^{it\lambda}-e^{is\lambda}|^2\,d\mu_x(\lambda). \end{equation*}

As \(t\to s\text{,}\) the integrand converges pointwise to zero. Moreover,

\begin{equation*} |e^{it\lambda}-e^{is\lambda}|^2\le 4. \end{equation*}

Since \(\mu_x(\mathbb R)=\|x\|^2<\infty\text{,}\) dominated convergence yields

\begin{equation*} \|U(t)x-U(s)x\|\to 0. \end{equation*}

Thus \(t\mapsto U(t)x\) is continuous for every \(x\in H\text{,}\) and the group is strongly continuous.

Remark 9.6.4.

This argument proves the Hille--Yosida theorem only in the self-adjoint Hilbert-space case. The full Banach-space Hille--Yosida theorem requires different methods, such as Yosida approximants, and does not follow directly from the spectral theorem.