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For an integer $s\ge 0$, the Sobolev space iswhere $D^{\beta }$ is a weak derivative. For $p<\infty$ one may use the Sobolev norm $({\sum }_{|\beta |\le s}\Vert D^{\beta }u{\Vert }_p^p{)}^{1/p}$; for $p=\infty$ use the maximum of the finitely many essential-supremum norms.

For $1\le p<d$, the Sobolev inequality is $\Vert u{\Vert }_{p^{\ast }}\le C_{d,p}\Vert \nabla u{\Vert }_p$, with Sobolev conjugate exponent $p^{\ast }=dp/(d-p)$. For $d<p<\infty$, Morrey's inequality supplies a continuous representative satisfyingAt $p=\infty$ the representative is Lipschitz continuous. At the critical exponent $p=d>1$, first-order Sobolev regularity gives every finite $L^q$ embedding for $q\ge d$, with an inhomogeneous norm, but generally no $L^{\infty }$ embedding. The one-dimensional endpoint $W^{1,1}(\mathbb{R})\hookrightarrow L^{\infty }(\mathbb{R})$ is an exception.

For the proof of Morrey's inequality, start with a smooth $u$ and write $u_{B(x,r)}$ for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables givesThe last step is the Holder inequality; integrability of the kernel to power $p^{\prime }$ is exactly $(d-1)p^{\prime }<d$. For $r=|x-y|$, translate the averaging ball along the segment from $x$ to $y$. The fundamental theorem of calculus along a line segment and the Holder inequality giveCombining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with $r=1$, together with $|u_{B(x,1)}|\le |B_1{|}^{-1/p}\Vert u{\Vert }_p$, gives the supremum estimate. Density of smooth functions in a Sobolev space then gives a uniformly convergent sequence of smooth representatives, preserving both bounds. For $p=\infty$, mollification gives the Lipschitz version.

For the decay conclusion assume $d<p<\infty$. The representative is uniformly continuous. If $|u(x_j)|\ge 2\theta >0$ along points escaping to infinity, the Hölder bound gives a radius $\rho >0$, independent of $j$, on which $|u|\ge \theta$. A subsequence has disjoint radius-$\rho$ balls, each contributing at least ${\theta }^p|B_{\rho }|$ to $\Vert u{\Vert }_p^p$, a contradiction. This is uniformly continuous integrable functions vanish at infinity.

The finite-$p$ restriction is necessary. If the printed range includes $p=\infty$, its decay assertion is false: $u\equiv 1$ belongs to $W^{1,\infty }$ but does not tend to zero.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

For $1\le p<n$, define the Sobolev conjugate exponentThe Sobolev inequality, also called the Gagliardo--Nirenberg--Sobolev inequality, states that there is a constant $C=C(n,p)$ such thatfor every $u\in C_c^{\infty }({\mathbb{R}}^n)$, and hence by completion for every $u\in W^{1,p}({\mathbb{R}}^n)$ for which the right formulation applies.