[proofplan] We write the heat semigroup as convolution with the heat kernel and move the spatial derivative onto the kernel. The derivative of the Gaussian has an exact parabolic scaling, so its $L^r$ norm has the factor $t^{-k/2-n(1-1/r)/2}$. Choosing $r$ by the Young convolution relation $1+1/q=1/p+1/r$ gives exactly the exponent in the desired estimate. The argument first identifies the derivative at the kernel level and then applies [Young's convolution inequality](/theorems/463) to extend the estimate to all $L^p$ data. [/proofplan] [step:Differentiate the heat kernel convolution in the spatial variable] Let $\mathcal{L}^n$ denote [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and let $C_c^\infty(\mathbb{R}^n)$ denote the space of smooth functions $\mathbb{R}^n\to\mathbb{R}$ with compact support. For each $t>0$, define the heat kernel $\Gamma_t: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} \Gamma_t(x)=(4\pi t)^{-n/2} e^{-\frac{|x|^2}{4t}}. \end{align*} For $f \in L^p(\mathbb{R}^n)$, define the heat semigroup function $e^{t\Delta}f: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} (e^{t\Delta}f)(x)=\int_{\mathbb{R}^n} \Gamma_t(x-y) f(y)\, d\mathcal{L}^n(y). \end{align*} For a multi-index $\alpha \in \mathbb{N}_0^n$ with $|\alpha|=k$, define the derivative kernel $K_{t,\alpha}: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} K_{t,\alpha}(x)=D^\alpha \Gamma_t(x). \end{align*} Since $\Gamma_t$ is smooth and every derivative of $\Gamma_t$ is a polynomial multiple of $\Gamma_t$, $K_{t,\alpha} \in L^r(\mathbb{R}^n)$ for every $1 \leq r \leq \infty$. Let $p'\in[1,\infty]$ denote the Hölder conjugate exponent of $p$, with $1/p+1/p'=1$. For each multi-index $\beta\in\mathbb{N}_0^n$ with $|\beta|\leq k$, the function $D^\beta\Gamma_t$ lies in $L^{p'}(\mathbb{R}^n)$, so Hölder's inequality shows that the integral defining $(D^\beta\Gamma_t)*f$ is absolutely convergent for each $x\in\mathbb{R}^n$. The Gaussian derivative bound also gives an $L^{p'}$ dominating function in a neighbourhood of each $x$, so differentiation under the integral is justified for each derivative up to order $k$. Therefore \begin{align*} D^\alpha e^{t\Delta}f=K_{t,\alpha}*f. \end{align*} [guided] We want to estimate a derivative of $e^{t\Delta}f$, and the useful point is that the heat semigroup is a convolution operator. Let $\mathcal{L}^n$ denote Lebesgue measure on $\mathbb{R}^n$, and let $C_c^\infty(\mathbb{R}^n)$ denote the space of smooth functions $\mathbb{R}^n\to\mathbb{R}$ with compact support. For each $t>0$, the heat kernel is the smooth function $\Gamma_t: \mathbb{R}^n \to \mathbb{R}$ defined by \begin{align*} \Gamma_t(x)=(4\pi t)^{-n/2} e^{-\frac{|x|^2}{4t}}. \end{align*} Thus the heat semigroup applied to $f \in L^p(\mathbb{R}^n)$ is the function $e^{t\Delta}f: \mathbb{R}^n \to \mathbb{R}$ defined by \begin{align*} (e^{t\Delta}f)(x)=\int_{\mathbb{R}^n} \Gamma_t(x-y) f(y)\, d\mathcal{L}^n(y). \end{align*} Now fix a multi-index $\alpha \in \mathbb{N}_0^n$ with $|\alpha|=k$. We introduce the derivative kernel $K_{t,\alpha}: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} K_{t,\alpha}(x)=D^\alpha \Gamma_t(x). \end{align*} The reason for this definition is that differentiating the convolution in the $x$ variable differentiates only the factor $\Gamma_t(x-y)$, not the rough function $f(y)$. Since $\Gamma_t$ is a Gaussian, every spatial derivative $D^\alpha \Gamma_t$ is a polynomial in $x/\sqrt{t}$ multiplied by $\Gamma_t$. Hence $K_{t,\alpha}$ belongs to $L^r(\mathbb{R}^n)$ for every $1 \leq r \leq \infty$. Let $p'\in[1,\infty]$ denote the Hölder conjugate exponent of $p$, with $1/p+1/p'=1$. For each multi-index $\beta\in\mathbb{N}_0^n$ with $|\beta|\leq k$, the function $D^\beta\Gamma_t$ lies in $L^{p'}(\mathbb{R}^n)$. Hence Hölder's inequality gives absolute convergence of \begin{align*} \int_{\mathbb{R}^n} D^\beta\Gamma_t(x-y) f(y)\, d\mathcal{L}^n(y) \end{align*} for every $x\in\mathbb{R}^n$. Why is it legitimate to differentiate under the integral? In a small neighbourhood of a fixed point $x$, each derivative of $\Gamma_t(x-y)$ up to order $k$ is bounded by an $L^{p'}$ Gaussian derivative majorant in the $y$ variable; multiplying by $f\in L^p(\mathbb{R}^n)$ gives an integrable majorant by Hölder's inequality. The [dominated convergence theorem](/theorems/4) for difference quotients therefore permits differentiation under the integral, one coordinate derivative at a time, up to total order $k$. Consequently \begin{align*} D^\alpha e^{t\Delta}f(x)=\int_{\mathbb{R}^n} D^\alpha \Gamma_t(x-y) f(y)\, d\mathcal{L}^n(y)=(K_{t,\alpha}*f)(x). \end{align*} [/guided] [/step] [step:Compute the parabolic scaling of the derivative kernel] Let $\Gamma_1: \mathbb{R}^n \to \mathbb{R}$ be defined by \begin{align*} \Gamma_1(z)=(4\pi)^{-n/2} e^{-\frac{|z|^2}{4}}. \end{align*} The scaling identity for the heat kernel is \begin{align*} \Gamma_t(x)=t^{-n/2}\Gamma_1(t^{-1/2}x). \end{align*} Differentiating this identity gives \begin{align*} D^\alpha \Gamma_t(x)=t^{-n/2-k/2}(D^\alpha\Gamma_1)(t^{-1/2}x). \end{align*} For $1 \leq r < \infty$, the change of variables $z=t^{-1/2}x$, equivalently $x=t^{1/2}z$, transforms Lebesgue measure by \begin{align*} d\mathcal{L}^n(x)=t^{n/2}\,d\mathcal{L}^n(z). \end{align*} Therefore \begin{align*} \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)} = t^{-n/2-k/2+n/(2r)} \|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)}. \end{align*} For $r=\infty$, taking the essential supremum in the same scaling identity gives \begin{align*} \|K_{t,\alpha}\|_{L^\infty(\mathbb{R}^n)} = t^{-n/2-k/2} \|D^\alpha\Gamma_1\|_{L^\infty(\mathbb{R}^n)}. \end{align*} Thus, for every $1 \leq r \leq \infty$, there is a finite constant \begin{align*} A_{\alpha,r}:=\|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)} \end{align*} such that \begin{align*} \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)} \leq A_{\alpha,r} t^{-k/2-\frac{n}{2}\left(1-\frac{1}{r}\right)}. \end{align*} [guided] The heat kernel has parabolic scaling: space scales like $t^{1/2}$, while the prefactor scales like $t^{-n/2}$. Define $\Gamma_1: \mathbb{R}^n\to\mathbb{R}$ by \begin{align*} \Gamma_1(z)=(4\pi)^{-n/2}e^{-\frac{|z|^2}{4}}. \end{align*} Then \begin{align*} \Gamma_t(x)=t^{-n/2}\Gamma_1(t^{-1/2}x). \end{align*} Applying the chain rule once contributes one factor $t^{-1/2}$; applying it $k=|\alpha|$ times gives \begin{align*} D^\alpha\Gamma_t(x)=t^{-n/2-k/2}(D^\alpha\Gamma_1)(t^{-1/2}x). \end{align*} For $1\leq r<\infty$, we compute the $L^r$ norm using the change of variables $z=t^{-1/2}x$, equivalently $x=t^{1/2}z$. Under this substitution, Lebesgue measure transforms as \begin{align*} d\mathcal{L}^n(x)=t^{n/2}\,d\mathcal{L}^n(z). \end{align*} Therefore \begin{align*} \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)}=t^{-n/2-k/2+n/(2r)}\|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)}. \end{align*} For $r=\infty$, the same scaling identity gives \begin{align*} \|K_{t,\alpha}\|_{L^\infty(\mathbb{R}^n)}=t^{-n/2-k/2}\|D^\alpha\Gamma_1\|_{L^\infty(\mathbb{R}^n)}. \end{align*} Thus for every $1\leq r\leq\infty$, with \begin{align*} A_{\alpha,r}:=\|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)}, \end{align*} we have \begin{align*} \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)}\leq A_{\alpha,r}t^{-k/2-\frac{n}{2}\left(1-\frac{1}{r}\right)}. \end{align*} [/guided] [/step] [step:Choose the Young exponent matching the desired smoothing] Define $r \in [1,\infty]$ by \begin{align*} \frac{1}{r}=1+\frac{1}{q}-\frac{1}{p}. \end{align*} Because $1 \leq p \leq q \leq \infty$, we have $0 \leq 1/r \leq 1$, so this definition is valid. Moreover, \begin{align*} 1+\frac{1}{q}=\frac{1}{p}+\frac{1}{r}. \end{align*} The standard Young convolution inequality for this triple of exponents gives \begin{align*} \|K_{t,\alpha}*f\|_{L^q(\mathbb{R}^n)} \leq \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)} \|f\|_{L^p(\mathbb{R}^n)}. \end{align*} [guided] We now choose the exponent that makes Young's convolution inequality produce exactly the target $L^p\to L^q$ estimate. Define $r\in[1,\infty]$ by \begin{align*} \frac{1}{r}=1+\frac{1}{q}-\frac{1}{p}. \end{align*} Because $1\leq p\leq q\leq\infty$, we have $1/p\geq 1/q$, so $1/r\leq 1$. Also $1/p-1/q\leq 1$, so $1/r\geq 0$. Hence this formula defines a valid exponent $r\in[1,\infty]$. Rearranging the definition gives the Young relation \begin{align*} 1+\frac{1}{q}=\frac{1}{p}+\frac{1}{r}. \end{align*} Young's convolution inequality applies to $K_{t,\alpha}\in L^r(\mathbb{R}^n)$ and $f\in L^p(\mathbb{R}^n)$ with this triple of exponents, giving \begin{align*} \|K_{t,\alpha}*f\|_{L^q(\mathbb{R}^n)}\leq \|K_{t,\alpha}\|_{L^r(\mathbb{R}^n)}\|f\|_{L^p(\mathbb{R}^n)}. \end{align*} This is the point where the condition $p\leq q$ is used: it ensures that the exponent $r$ required by convolution is not outside the allowed range. [/guided] [/step] [step:Substitute the kernel norm and simplify the exponent] Combining the identity $D^\alpha e^{t\Delta}f=K_{t,\alpha}*f$ with the preceding convolution estimate gives \begin{align*} \|D^\alpha e^{t\Delta}f\|_{L^q(\mathbb{R}^n)} \leq A_{\alpha,r} t^{-k/2-\frac{n}{2}\left(1-\frac{1}{r}\right)} \|f\|_{L^p(\mathbb{R}^n)}. \end{align*} From the definition of $r$, \begin{align*} 1-\frac{1}{r} = 1-\left(1+\frac{1}{q}-\frac{1}{p}\right) = \frac{1}{p}-\frac{1}{q}. \end{align*} Therefore \begin{align*} \|D^\alpha e^{t\Delta}f\|_{L^q(\mathbb{R}^n)} \leq A_{\alpha,r} t^{-\frac{k}{2}-\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)} \|f\|_{L^p(\mathbb{R}^n)}. \end{align*} Since there are only finitely many multi-indices $\alpha \in \mathbb{N}_0^n$ with $|\alpha|=k$, define \begin{align*} C(n,k,p,q):= \max_{|\alpha|=k} \|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)}. \end{align*} This constant is finite because each $D^\alpha\Gamma_1$ is a polynomial multiple of a Gaussian. The desired estimate follows for every multi-index $\alpha$ with $|\alpha|=k$. [guided] We combine the pieces. From the first step, \begin{align*} D^\alpha e^{t\Delta}f=K_{t,\alpha}*f. \end{align*} [Young's inequality](/theorems/244) and the kernel estimate therefore give \begin{align*} \|D^\alpha e^{t\Delta}f\|_{L^q(\mathbb{R}^n)}\leq A_{\alpha,r}t^{-k/2-\frac{n}{2}\left(1-\frac{1}{r}\right)}\|f\|_{L^p(\mathbb{R}^n)}. \end{align*} The exponent now simplifies because the definition of $r$ gives \begin{align*} 1-\frac{1}{r}=\frac{1}{p}-\frac{1}{q}. \end{align*} Substituting this identity yields \begin{align*} \|D^\alpha e^{t\Delta}f\|_{L^q(\mathbb{R}^n)}\leq A_{\alpha,r}t^{-\frac{k}{2}-\frac{n}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|f\|_{L^p(\mathbb{R}^n)}. \end{align*} The theorem asks for one constant that works for every multi-index of order $k$, not a different constant for each $\alpha$. The set of multi-indices $\alpha\in\mathbb{N}_0^n$ with $|\alpha|=k$ is finite, so we may define \begin{align*} C(n,k,p,q):=\max_{|\alpha|=k}\|D^\alpha\Gamma_1\|_{L^r(\mathbb{R}^n)}. \end{align*} This maximum is finite because every derivative $D^\alpha\Gamma_1$ is a polynomial multiple of a Gaussian, hence belongs to $L^r(\mathbb{R}^n)$ for every $1\leq r\leq\infty$. Replacing $A_{\alpha,r}$ by this maximum gives the asserted estimate uniformly for all $|\alpha|=k$. [/guided] [/step]