2.3 The Heat Semigroup

Definition 2.3.1 (Normal Operator).label Let $H$ be a Hilbert space and $T$ be a closed, densely defined operator, then $T$ is normal if $T^{*}T = TT^{*}$.

Theorem 2.3.2 (Spectral Theorem I, [The Spectral Theorem, Con85]).label Let $H$ be a Hilbert space and $T$ be a normal operator, then there exists a unique spectral measure $P: \cb_{\complex} \to L(H; H)$ such that:

  1. (1)

    $T = \int z P(dz)$.

  2. (2)

    The mapping $f \mapsto \int f(z) P(dz)$ is a $*$-homomorphism.

Definition 2.3.3.label Let $L$ be an essentially self-adjoint diffusion operator and $P$ be the spectral measure associated with $-L$. For each $t \ge 0$, let

\[\bp_{t} = \int_{\complex} e^{-tz}P(dz)\]

then:

  1. (1)

    For any $f \in L^{2}(\real^{d})$, $\norm{\bp_tf}_{L^2(\real^d, \mu)}\le \norm{f}_{L^2(\real^d, \mu)}$.

  2. (2)

    For each $t \ge 0$, $\bp_{t}$ is self-adjoint.

  3. (3)

    $\bracs{\bp_t|t \ge 0}$ is a $C_{0}$-semigroup with $L$ as its generator.

The family $\bracs{\bp_t|t \ge 0}$ is the heat semigroup associated with $L$.

Proposition 2.3.4.label Let $L$ be an essentially self-adjoint elliptic diffusion operator with smooth coefficients, and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup, then:

  1. (1)

    For any precompact open set $U \subset \subset \real^{d}$ and $k \in \nat$,

    \[\norm{P_tf}_{H^k(K)}\le C_{K, k, \mu}\paren{1 + \frac{1}{t^{k}}}\norm{f}_{L^2(\real^d)}\]

    In particular, by the Sobolev Embedding theorem,

    \[\sup_{x \in K}|\bp_{t}f(x)| \le C_{K}\paren{1 + \frac{1}{t^{\lfloor d/2 \rfloor + 1}}}\norm{f}_{L^2(\real^d)}\]

  2. (2)

    For any $f \in L^{2}(\real^{d}; \real)$, the mapping $(t, x) \mapsto \bp_{t}f(x)$ is smooth on $(0, \infty) \times \real^{d}$.

Proof. (1): For any $k \in \nat$,

\[L^{k}\bp_{t} = \int z^{k} e^{-tz}P(dz)\]

Hence

\[\normn{L^k \bp_t}_{L(L^2(\mu), L^2(\mu))}\le \sup_{z \ge 0}z^{k}e^{-z t}\le \frac{k^{k}}{t^{k}}e^{-k}\le \frac{C_{k}}{t^{k}}\]

By the Elliptic Regularity Theorem,

\begin{align*}\norm{P_tf}_{H^k(K)}&\le C_{K}\sum_{\ell = 0}^{k} \normn{L^k\bp_tf}_{L^2(\real^d)}\\&\le C_{K, k, \mu}\paren{1 + \frac{1}{t^{k}}}\norm{f}_{L^2(\real^d)}\end{align*}

(2): By (1), $\bp_{t} f \in H^{k}_{\text{loc}}$ for all $k \in \nat$, so $\bp_{t} f \in C^{\infty}$. For any $s, t > 0$ and $k \in \nat$,

\[\norm{L^k\bp_t - L^k\bp_s}_{L(L^2(\mu), L^2(\mu))}\le \sup_{z \ge 0}|{z^ke^{-z}(e^{-t} - e^{-s})}|\]

so $(t, x) \mapsto \bp_{t} f(x)$ is jointly continuous. By a very wasteful chain of Sobolev shenanigans, $(t, x) \mapsto \bp_{t} f(x)$ is smooth.$\square$

Theorem 2.3.5.label Let $L$ be an essentially self-adjoint elliptic diffusion operator and $\bracs{\bp_t|t \ge 0}$ be its heat semigroup, then there exists $p \in C^{\infty}(\real \times \real^{d} \times \real^{d}; \real)$ such that

  1. (1)

    For any $f \in L^{2}(\real^{d}; \real)$ and $x \in \real^{d}$,

    \[\bp_{t} f(x) = \int p(t, x, y)f(y) \mu(dy)\]

  2. (2)

    For any $x, y \in \real^{d}$ and $t > 0$, $p(t, x, y) = p(t, y, x)$.

  3. (3)

    Chapman-Kolmogorov Relation: For any $x, y \in \real^{d}$ and $s, t > 0$,

    \[p(s + t, x, y) = \int p(t, x, z)p(s, z, y)\mu(dz)\]

Proof. (1): For each $t > 0$ and $x \in \real^{d}$, the mapping $f \mapsto \bp_{t}f(x)$ is a continuous linear functional on $L^{2}(\mu)$. Hence there exists a continuous mapping

\[p: \real \times \real^{d} \to L^{2}(\mu)\]

such that

\[\bp_{t} f(x) = \int p(t, x, y)f(y)\mu(dy)\]

(2): Since $\bp_{t}$ is self-adjoint on $L^{2}(\mu)$, $p$ is symmetric.

(3): Since $\bracs{\bp_t|t \ge 0}$ is a semigroup, $p$ satisfies the Chapman-Kolmogorov relation.

Smoothness: For any $f \in L^{2}(\mu)$,

\[(t, x) \mapsto \int p(t, x, y) f(y)dy\]

is smooth. Thus, viewed as a map $\real \times \real^{d} \to L^{2}(\mu)$, $(t, x) \mapsto p(t, x, \cdot)$ is differentiable with respect to the weak topology on $L^{2}(\mu)$.

Since the inner product $L^{2}(\mu) \times L^{2}(\mu) \to \real$ is smooth, by the Chapman-Kolmogorov relation, for any $t > 0$,

\[p(t, x, y) = \int p(t/2, x, z)p(t/2, z, y)\mu(dz)\]

is continuous.$\square$