[proofplan]
We represent the heat semigroup as convolution with the heat kernel $\Gamma_t$. The exponent $r$ is chosen so that [Young's convolution inequality](/theorems/463) maps $L^r * L^p$ into $L^q$. The only remaining work is to compute the $L^r$ norm of $\Gamma_t$ by the parabolic scaling $x = \sqrt{t}\,y$, including the endpoint case $r=\infty$. The relation among $p,q,r$ then converts the scaling exponent into $\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)$.
[/proofplan]
[step:Choose the Young exponent connecting $L^p$ to $L^q$]
Use the convention $1/\infty := 0$. Define $r \in [1,\infty]$ by
\begin{align*}
\frac{1}{r} := 1 + \frac{1}{q} - \frac{1}{p}.
\end{align*}
Because $p \leq q$, we have $\frac{1}{p} \geq \frac{1}{q}$, so $\frac{1}{r} \leq 1$. Also $\frac{1}{p} \leq 1$ and $\frac{1}{q} \geq 0$, so $\frac{1}{r} \geq 0$. Hence $r \in [1,\infty]$ is well-defined, and it satisfies
\begin{align*}
1 + \frac{1}{q} = \frac{1}{p} + \frac{1}{r}.
\end{align*}
[/step]
[step:Compute the $L^r$ norm of the heat kernel]
Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$. For each $t > 0$, define the heat kernel $\Gamma_t: \mathbb{R}^n \to \mathbb{R}$ by the rule
\begin{align*}
\Gamma_t(x) := (4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right) \text{ for } x \in \mathbb{R}^n.
\end{align*}
We claim that there is a constant $A_{n,r} > 0$ such that
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)} = A_{n,r}t^{-\frac{n}{2}\left(1-\frac{1}{r}\right)}
\end{align*}
for every $t > 0$, with the interpretation $1/r=0$ when $r=\infty$.
If $1 \leq r < \infty$, then
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)}^r = \int_{\mathbb{R}^n} (4\pi t)^{-nr/2}\exp\left(-\frac{r|x|^2}{4t}\right)\,d\mathcal{L}^n(x).
\end{align*}
Apply the change of variables $x=\sqrt{t}\,y$. Under this substitution, $d\mathcal{L}^n(x)=t^{n/2}\,d\mathcal{L}^n(y)$ and the domain $\mathbb{R}^n$ is mapped onto $\mathbb{R}^n$. Therefore
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)}^r = (4\pi)^{-nr/2}t^{-nr/2+n/2}\int_{\mathbb{R}^n}\exp\left(-\frac{r|y|^2}{4}\right)\,d\mathcal{L}^n(y).
\end{align*}
Define
\begin{align*}
A_{n,r} := (4\pi)^{-n/2}\left(\int_{\mathbb{R}^n}\exp\left(-\frac{r|y|^2}{4}\right)\,d\mathcal{L}^n(y)\right)^{1/r}.
\end{align*}
The [Gaussian integral](/theorems/1140) is finite, so taking the $r$-th root gives
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)} = A_{n,r}t^{-\frac{n}{2}\left(1-\frac{1}{r}\right)}.
\end{align*}
If $r=\infty$, then $\Gamma_t$ is maximized at $x=0$, since $\exp(-|x|^2/(4t)) \leq 1$ for every $x \in \mathbb{R}^n$. Hence
\begin{align*}
\|\Gamma_t\|_{L^\infty(\mathbb{R}^n)} = (4\pi t)^{-n/2} = (4\pi)^{-n/2}t^{-n/2}.
\end{align*}
This is the same formula with $A_{n,\infty}:=(4\pi)^{-n/2}$ and $1/r=0$.
[guided]
The heat kernel has exactly the scaling expected from the [heat equation](/page/Heat%20Equation): space scales like $\sqrt{t}$. We verify this directly because the whole theorem is driven by this scaling factor.
For $1 \leq r < \infty$, the $L^r(\mathbb{R}^n)$ norm is defined by integration with respect to $n$-dimensional Lebesgue measure:
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)}^r = \int_{\mathbb{R}^n} |\Gamma_t(x)|^r\,d\mathcal{L}^n(x).
\end{align*}
Since $\Gamma_t(x)>0$, substituting the definition of $\Gamma_t$ gives
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)}^r = \int_{\mathbb{R}^n} (4\pi t)^{-nr/2}\exp\left(-\frac{r|x|^2}{4t}\right)\,d\mathcal{L}^n(x).
\end{align*}
Now use the scaling substitution $x=\sqrt{t}\,y$, where $y \in \mathbb{R}^n$. The [linear map](/page/Linear%20Map) $y \mapsto \sqrt{t}\,y$ has determinant $t^{n/2}$, so Lebesgue measure transforms as
\begin{align*}
d\mathcal{L}^n(x)=t^{n/2}\,d\mathcal{L}^n(y).
\end{align*}
The substitution maps $\mathbb{R}^n$ onto $\mathbb{R}^n$, and $|x|^2=t|y|^2$. Therefore
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)}^r = (4\pi)^{-nr/2}t^{-nr/2+n/2}\int_{\mathbb{R}^n}\exp\left(-\frac{r|y|^2}{4}\right)\,d\mathcal{L}^n(y).
\end{align*}
The integral
\begin{align*}
\int_{\mathbb{R}^n}\exp\left(-\frac{r|y|^2}{4}\right)\,d\mathcal{L}^n(y)
\end{align*}
is a finite Gaussian integral because $r \geq 1$. Define the constant
\begin{align*}
A_{n,r} := (4\pi)^{-n/2}\left(\int_{\mathbb{R}^n}\exp\left(-\frac{r|y|^2}{4}\right)\,d\mathcal{L}^n(y)\right)^{1/r}.
\end{align*}
Taking the $r$-th root yields
\begin{align*}
\|\Gamma_t\|_{L^r(\mathbb{R}^n)} = A_{n,r}t^{-\frac{n}{2}\left(1-\frac{1}{r}\right)}.
\end{align*}
The endpoint $r=\infty$ is not obtained by taking an integral. Instead, the $L^\infty$ norm is the essential supremum. Since
\begin{align*}
\exp\left(-\frac{|x|^2}{4t}\right) \leq 1
\end{align*}
for every $x \in \mathbb{R}^n$, and equality holds at $x=0$, we get
\begin{align*}
\|\Gamma_t\|_{L^\infty(\mathbb{R}^n)} = (4\pi t)^{-n/2}.
\end{align*}
Thus the same scaling formula holds with $A_{n,\infty}:=(4\pi)^{-n/2}$ and $1/\infty=0$.
[/guided]
[/step]
[step:Apply Young's convolution inequality to the heat kernel and the data]
Let $f \in L^p(\mathbb{R}^n)$ and $t>0$. Since $\Gamma_t \in L^r(\mathbb{R}^n)$ by the preceding step and $f \in L^p(\mathbb{R}^n)$ by hypothesis, Young's convolution inequality applies to the pair
\begin{align*}
\Gamma_t \in L^r(\mathbb{R}^n), \qquad f \in L^p(\mathbb{R}^n),
\end{align*}
with exponents satisfying
\begin{align*}
1+\frac{1}{q}=\frac{1}{r}+\frac{1}{p}.
\end{align*}
Thus, for a constant $Y_{p,r,q}>0$ depending only on $p,r,q$,
\begin{align*}
\|\Gamma_t*f\|_{L^q(\mathbb{R}^n)} \leq Y_{p,r,q}\|\Gamma_t\|_{L^r(\mathbb{R}^n)}\|f\|_{L^p(\mathbb{R}^n)}.
\end{align*}
Here the convolution is the measurable function $\Gamma_t*f: \mathbb{R}^n \to \mathbb{R}$ given by
\begin{align*}
(\Gamma_t*f)(x) := \int_{\mathbb{R}^n}\Gamma_t(x-y)f(y)\,d\mathcal{L}^n(y), \qquad x \in \mathbb{R}^n,
\end{align*}
with the usual almost-everywhere interpretation when $f$ is an $L^p$ equivalence class.
Substituting the heat-kernel norm estimate gives
\begin{align*}
\|\Gamma_t*f\|_{L^q(\mathbb{R}^n)} \leq Y_{p,r,q}A_{n,r}t^{-\frac{n}{2}\left(1-\frac{1}{r}\right)}\|f\|_{L^p(\mathbb{R}^n)}.
\end{align*}
[/step]
[step:Rewrite the scaling exponent in terms of $p$ and $q$]
From the defining relation
\begin{align*}
\frac{1}{r}=1+\frac{1}{q}-\frac{1}{p},
\end{align*}
we obtain
\begin{align*}
1-\frac{1}{r}=\frac{1}{p}-\frac{1}{q}.
\end{align*}
Therefore
\begin{align*}
t^{-\frac{n}{2}\left(1-\frac{1}{r}\right)} = t^{-\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}.
\end{align*}
Define
\begin{align*}
C(n,p,q) := Y_{p,r,q}A_{n,r}.
\end{align*}
Since $r$ is determined by $p$ and $q$, this constant depends only on $n,p,q$. Hence
\begin{align*}
\|\Gamma_t*f\|_{L^q(\mathbb{R}^n)} \leq C(n,p,q)t^{-\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|f\|_{L^p(\mathbb{R}^n)}.
\end{align*}
By the definition $e^{t\Delta}f:=\Gamma_t*f$, this is precisely
\begin{align*}
\|e^{t\Delta}f\|_{L^q(\mathbb{R}^n)} \leq C(n,p,q)t^{-\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|f\|_{L^p(\mathbb{R}^n)}.
\end{align*}
This proves the estimate for every $t>0$ and every $f \in L^p(\mathbb{R}^n)$.
[/step]
