For a flux and a weak solution of , discrete integration by parts and the fundamental theorem of calculus along a line segment show that solves , with and . Coercive quadratic-form bounds and operator norm bounds for on all segments between these gradients pass to . This produces a divergence-form elliptic operator without initially assuming second partial derivatives of .
For a cube of side length in , , and a continuously differentiable function , its average satisfies
Averaging the fundamental theorem of calculus along a line segment from to and changing variables gives . The Holder inequality applies because , and the kernel norm is . Approximation extends the estimate to the continuous representative of a Sobolev space function. On it yields a Hölder continuous function of exponent representing every class.
Morrey's inequality 2026-10-06
For , every Sobolev space element has a Hölder continuous function representative of exponent , with the displayed bound. It also satisfies . Averaging the fundamental theorem of calculus along a line segment over a ball gives
The Holder inequality bounds this by , because . To compare two ball averages, translate a ball along the segment between their centres and use the same fundamental theorem of calculus along a line segment. These bounds give the displayed estimate. Density of smooth functions in a Sobolev space supplies the representative for nonsmooth . For , the corresponding conclusion is Lipschitz continuity.
For an integer , the Sobolev space is
where is a weak derivative. For one may use the Sobolev norm ; for use the maximum of the finitely many essential-supremum norms.
For , the Sobolev inequality is , with Sobolev conjugate exponent . For , Morrey's inequality supplies a continuous representative satisfying
At the representative is Lipschitz continuous. At the critical exponent , first-order Sobolev regularity gives every finite embedding for , with an inhomogeneous norm, but generally no embedding. The one-dimensional endpoint is an exception.
For the proof of Morrey's inequality, start with a smooth and write for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables gives
The last step is the Holder inequality; integrability of the kernel to power is exactly . For , translate the averaging ball along the segment from to . The fundamental theorem of calculus along a line segment and the Holder inequality give
Combining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with , together with , 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 , mollification gives the Lipschitz version.
For the decay conclusion assume . The representative is uniformly continuous. If along points escaping to infinity, the Hölder bound gives a radius , independent of , on which . A subsequence has disjoint radius- balls, each contributing at least to , a contradiction. This is uniformly continuous integrable functions vanish at infinity.
The finite- restriction is necessary. If the printed range includes , its decay assertion is false: belongs to but does not tend to zero.
Fix nested relatively compact subsets of and let bound the gradient on the larger one. The fundamental theorem of calculus along a line segment gives on the smaller one. Part (i) gives coefficient bounds independent of small , so the De Giorgi-Nash-Moser theorem gives uniform interior Hölder continuity estimates for these weak solutions. Since is continuous, the difference quotients converge locally uniformly to . Passing to the limit in the Hölder seminorm estimate, and repeating for every coordinate, yields
The first derivatives are locally Hölder continuous. Here the initially obtained exponent depends on and the local gradient bound. This is the interior interpretation of the printed regularity conclusion; a uniform global estimate does not follow from alone.
Write the minimal surface equation for a graph as in its weak formulation, where . For negative increments use the same difference quotient convention, so
For , the function is an admissible test function in . Commuting the first partial derivative with translation, then applying discrete integration by parts, gives
The fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equation
Thus the backward difference quotient solves a linear equation in divergence form:
To verify the ellipticity of the minimal surface flux, compute
and hence
The eigenvalue parallel to is ; every orthogonal eigenvalue is . The same lower and upper bounds pass to the averaged symmetric matrix whenever both endpoint gradients have norm at most :
The printed hypothesis supplies such an on each relatively compact subset, uniformly for sufficiently small . It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of additionally requires bounded gradients there and at the shifted points; this is automatic for bounded and fixed , but is not supplied on an arbitrary open set.
For distinct , convexity keeps the segment in . Continuity of makes
The fundamental theorem of calculus along a line segment gives
so . Thus is injective.
The bound makes invertible by the Neumann series. The inverse function theorem therefore makes locally open at every point, so is open.
Surjectivity need not hold: the identity map on a proper convex open set has proper image. Even for it can fail. In one dimension, satisfies , but its image is .
Put and . The fundamental theorem of calculus along a line segment, followed by averaging, gives
For fixed , use the change of variables formula . Its Jacobian determinant is . The transformed domain lies in , because a cube is a convex set. Moreover, implies . Thus Tonelli theorem gives
The diagonal is a null set. The Holder inequality with now applies: implies , and integration in spherical coordinates bounds the kernel by
This proves the Morrey inequality on a cube, uniformly even when approaches its boundary:
Apply this at both and the center and use the triangle inequality to obtain