Strong continuity of the heat semigroup in $L^2(\mathbb{R}^n)$. We know that $(K_t * g)(x) = (4\pi t)^{-n/2} \int_{\mathbb{R}^n} e^{-||x-y||^2/4t}g(y)dy$ solves the heat equation $u_t-\Delta u = 0$ in $\mathbb{R}^n \times (0 , \infty)$ with $u=g$ in $\mathbb{R}^n \times \{0\}$. I know how to estimate $\lim_{t \rightarrow 0^{+}}|| (K_t * g)(x) - g(x) ||=0$ in $L^1(\mathbb{R}^n) $ and $L^{\infty}(\mathbb{R}^n) $.
My question is, how to extend this to $L^2(\mathbb{R}^n) $?
Is there any interpolation inequality that does this directly? because when I try to do it directly I get a square integral that I try to use Holder’s inequality on and I don’t get anywhere:
$$ || (K_t * g)(x) - g(x) ||_2 = \left( \int_{\mathbb{R}^n} \left( (4\pi t)^{-n/2} \int_{\mathbb{R}^n} e^{-||x-y||^2/4t}g(y)dy -g(x) \right)^2 dx \right)^{1/2} = \left( \int_{\mathbb{R}^n} \left( (4\pi t)^{-n/2} \int_{\mathbb{R}^n} e^{-||x-y||^2/4t}[g(y)-g(x)]dy  \right)^2 dx \right)^{1/2}$$
above, I used that $ \int_{\mathbb{R}^n}(4\pi t)^{-n/2} e^{-||x-y||^2/4t}dy = 1.$
 A: Suppose first that $\frac{1}{p}+\frac{1}{q}=1$, and that $f\in L^p$. Then,
$$
          \left|\int K(x,y,t)f(y)dy\right|\le\int K(x,y,t)^{\frac{1}{p}}K(x,y,t)^{\frac{1}{q}}|f|dx \\
     \le\left(\int K(x,y,t)|f|^pdx\right)^{1/p}\left(\int K(x,y,t)dx\right)^{1/q} \\
      =\left(\int K(x,y,t)|f|^pdx\right)^{1/p}
$$
Now take the $p$-th power and integrate in $x$:
\begin{align}
         \int\left|\int K(x,y,t)f(y)dy\right|^pdx & \le \int\int K(x,y,t)|f|^pdx dy \\
           &= \int\int K(x,y,t)dy |f(x)|^pdx \\
           &=\int |f(x)|^pdx.
\end{align}
That gives the operator estimate $\|K_tf\|_p \le \|f\|_p$ for all $f\in L^p$, $1 < p < \infty$. The case for $p=1$ is handled separately in a straightforward manner. Therefore, $\|K_tf\|_{L^p}\le \|f\|_{L^p}$ for $1 \le p < \infty$.
All you have to do in order to show $\|K_tf-f\|_{L^p}\rightarrow 0$ as $t\downarrow 0$ for all $f \in L^p$, $1 \le p < \infty$ is to prove the result for a dense subspace of $L^p$, such as compactly supported, infinitely differentiable functions. $L^{\infty}$ does not work out as well because $\int K_t(x,y)f(y)dy$ will be smooth in $x$ for $t > 0$, and can't converge uniformly to a general $L^{\infty}$ function as $t\downarrow 0$.
A: I think I understand what needs to be done. Using the minkowsky inequality eliminates the square of the integral from within.
