Complex-time heat operators form a bounded holomorphic semigroup

For $\mathrm{\mid }\arg z\mathrm{\mid }\le \sigma <\theta$, [F4] and the $L^1$ bound of [F1] give $\mathrm{\parallel }{H}_{z}f{\mathrm{\parallel }}_p=\mathrm{\parallel }{\mathrm{\Gamma }}_z\ast f{\mathrm{\parallel }}_p\le \mathrm{\parallel }{\mathrm{\Gamma }}_z{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p\le (\cos \sigma {)}^{-n\mathrm{/}2}\mathrm{\parallel }f{\mathrm{\parallel }}_p$, which is (i).

Fix $x\in {\mathbb{R}}^n$ and a real $w>0$. The map $z\mapsto ({\mathrm{\Gamma }}_z\ast {\mathrm{\Gamma }}_w)(x)$ is holomorphic on $S_{\theta }$: pairing the $L^1$ difference quotients of [F3] with the bounded function $y\mapsto {\mathrm{\Gamma }}_w(x-y)$ proves scalar differentiability. The map $z\mapsto {\mathrm{\Gamma }}_{z+w}(x)$ is holomorphic by the kernel definition and [F2]. They agree for real $z>0$ by [F4], so [F2] gives equality for every complex $z\in S_{\theta }$. Now fix such a complex $z$. The same argument with the bounded kernel ${\mathrm{\Gamma }}_z$ makes $w\mapsto ({\mathrm{\Gamma }}_z\ast {\mathrm{\Gamma }}_w)(x)$ holomorphic, and $w\mapsto {\mathrm{\Gamma }}_{z+w}(x)$ is holomorphic because addition stays in the sector. They agree for real $w>0$ by the first application, so the identity theorem gives ${\mathrm{\Gamma }}_z\ast {\mathrm{\Gamma }}_w={\mathrm{\Gamma }}_{z+w}$ for all $z,w\in S_{\theta }$. Every scalar convolution is absolutely convergent since one kernel is bounded and the other is integrable.

The nonnegative double integral $\iint \mathrm{\mid }{\mathrm{\Gamma }}_z(x-y){\mathrm{\Gamma }}_w(y-v)f(v)\mathrm{\mid }dvdy$ is finite for almost every $x$: its $L^p$ bound is at most $\mathrm{\parallel }{\mathrm{\Gamma }}_z{\mathrm{\parallel }}_1\mathrm{\parallel }{\mathrm{\Gamma }}_w{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_p$ by two applications of Young [F4], with the pointwise bound for $p=\mathrm{\infty }$. Thus Fubini (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability) and step 1.2 give $H_z{H}_{w}f=({\mathrm{\Gamma }}_z\ast {\mathrm{\Gamma }}_w)\ast f={\mathrm{\Gamma }}_{z+w}\ast f=H_{z+w}f$ as $L^p$ classes. This proves (ii), including both exponent endpoints.

By [F3] and [F4], $\parallel (H_{z+h}f-H_{z}f)\mathrm{/}h-({\mathrm{\partial }}_z{\mathrm{\Gamma }}_z)\ast f{\parallel }_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\parallel ({\mathrm{\Gamma }}_{z+h}-{\mathrm{\Gamma }}_z)\mathrm{/}h-{\mathrm{\partial }}_z{\mathrm{\Gamma }}_z{\parallel }_1\to 0$, so $z\mapsto H_{z}f$ is norm differentiable with the stated derivative; the convergence of the difference quotients in $L^1$ is uniform on compact subsets of $S_{\theta }$ by [F3], so the same estimate gives local convergence in operator norm on every proper subsector, i.e. $z\mapsto H_z$ is holomorphic in operator norm on $S_{\sigma }$, $\sigma <\theta$.

Steps 1.1, 2.1 and 2.2 establish the boundedness estimate (i), the semigroup law (ii) and the differentiability statement (iii) with its operator-norm consequence.