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Adopt the sign convention that a weak supersolution for $Lu=D_i(a^{ij}{D}_{j}u)$ satisfies $Lu\le 0$, or $\int a^{ij}{D}_{j}u{D}_i\phi \ge 0$ for every nonnegative compactly supported test function. For a nonnegative supersolution, the Weak Harnack inequality gives some $q>0$ and $C_H\ge 1$, depending only on $n,\lambda ,\Lambda$, such thatHere $|E{|}^{-1}{\int }_E$ denotes the average over $E$; equivalently the right side gives a lower bound for the essential infimum. One may take a sufficiently small positive exponent, so no assumption $q\ge 1$ is needed. The inequality is unchanged under translations and dilations of the Euclidean balls.

Since a Sobolev space element is defined up to changes on a null set, the infimum is initially essential. After choosing a continuous representative it is the ordinary infimum. Without choosing such a representative, an arbitrary pointwise infimum cannot be bounded by an integral.

Work first with essential bounds. On a Euclidean ball $B_R(x_0)⋐B_1$, put $M={\mathrm{esssup}}_{B_R}u$, $m={\mathrm{essinf}}_{B_R}u$ and $\omega =M-m$. The functions $M-u$ and $u-m$ are nonnegative weak solutions, hence weak supersolutions. In $B_{R/2}$, at least one of the sets $\{u\le (M+m)/2\}$ and $\{u\ge (M+m)/2\}$ has at least half the measure.

In the first case, the Weak Harnack inequality applied to $M-u$ givesIn the second case, applying it to $u-m$ gives the same lower bound for the improvement of the minimum. Thus the oscillation decay estimate isThis measure argument works even when $q<1$, avoiding a false use of the triangle inequality in that range.

Iteration yields ${\mathrm{osc}}_{B_r(x_0)}u\le C(r/R{)}^{\mu }{\mathrm{osc}}_{B_R(x_0)}u$, with, for example, $\mu =\min \{1/2,-\log \tau /\log 2\}\in (0,1)$. At Lebesgue points the shrinking oscillations define a unique continuous representative, and the estimate extends to every point by continuity. For $x,y\in B_{1/4}$ at distance less than $1/4$, apply the estimate to a Euclidean ball centred at $x$ with initial radius $1/2$ and final radius comparable to $|x-y|$. This Euclidean ball stays in $B_1$. Larger distances use the trivial bound $2\Vert u{\Vert }_{\infty }$. We obtainThe constants depend only on $n,\lambda ,\Lambda$. This derives the relevant conclusion of the De Giorgi-Nash-Moser theorem directly from part (a).

First prove the hinted Caccioppoli inequality. Testing the weak formulation with $u_k{\phi }^2$ is legitimate by approximation in the zero-boundary Sobolev space. Let ${\Lambda }_{\ast }=\mathrm{esssup}\Vert a(x){\Vert }_{\mathrm{op}}\le n\Lambda$. Thenby Young inequality. HenceNo derivative of the measurable coefficient matrix is taken.

Set $w_k=u_k/{\sup }_{B_1}|u_k|$. The denominator is finite and positive because $u_k$ is continuous on the compact closure and nonzero. Each $w_k$ solves the same linear equation and $|w_k|\le 1$. For every $\theta <1$, choose a cutoff function equal to one on $B_{\theta }$ and supported in a slightly larger interior Euclidean ball. The energy bound gives a uniform $W^{1,2}(B_{\theta })$ bound.

Take radii ${\theta }_j\uparrow 1$. Weak compactness and a diagonal subsequence argument produce one subsequence converging weakly in $W^{1,2}(B_{{\theta }_j})$ for every $j$, with consistent restrictions defining $v\in W_{\mathrm{loc}}^{1,2}(B_1)$. Part (b) and the Arzela-Ascoli theorem allow a further subsequence converging uniformly on $\overline{B_{1/4}}$. Its uniform limit agrees almost everywhere with the weak limit and supplies its continuous representative there.

For every smooth compactly supported test function, choose $j$ containing its support. Since $a^{ij}{D}_i\phi \in L^2$ and $D{w}_k$ converges weakly there,Thus the compactness of normalized weak elliptic solutions givesThe diagonal step is needed to retain the equation beyond the small Euclidean ball of uniform convergence. The limit need not be nonzero: for the Laplacian in dimension at least two, the normalized harmonic functions $\mathrm{Re}(x_1+i{x}_2{)}^k$ tend uniformly to zero on each smaller Euclidean ball.

For the divergence-form elliptic operator use the weak subsolution convention $\int a^{ij}{D}_{j}u{D}_i\phi \le 0$ for nonnegative test functions. Set $u_{\epsilon }=u+\epsilon$ and test with $\phi ={\eta }^2{u}_{\epsilon }^{\beta -1}$, where $\beta >1$. The Sobolev chain rule makes this test admissible; boundedness and the positive regularization control the derivative of its power. Ellipticity and the Cauchy-Schwarz inequality give, with $X=\int |Du{|}^2{u}_{\epsilon }^{\beta -2}{\eta }^2$ and $Y=\int u_{\epsilon }^{\beta }|D\eta {|}^2$,Division when $X>0$, and the trivial case $X=0$, establish the power Caccioppoli inequality. Letting $\epsilon \downarrow 0$ givesFor $1<\beta <2$ this limit uses the Fatou lemma; the gradient of a Sobolev function vanishes on a level set, so the weighted integrand is assigned zero on $\{u=0\}$. The same regularization justifies differentiating $u^{\beta /2}\eta$ in $W^{1,2}$.

The PDF bounds individual entries by $\Lambda$. Then ${\Lambda }_{\ast }\le n\Lambda$: dimensional factors are included in the ellipticity constants when the first estimate is written as $C(\lambda ,\Lambda )$. With literal entrywise bounds, the explicit valid constant displayed here is $4n^2{\Lambda }^2/{\lambda }^2$. If $\Lambda$ bounds the operator norm instead, it is $4{\Lambda }^2/{\lambda }^2$. This distinction does not alter any later conclusion, whose constants explicitly depend on $n$.

For $n\ge 3$, put $\sigma =n/(n-2)$. Apply the Sobolev inequality to $g=u^{\beta /2}\eta$ and useFor $\beta \ge p>1$, $\beta /(\beta -1)\le p/(p-1)$, so the resulting constant is uniform in $\beta$. Taking a cutoff equal to one on $B_r$, supported in $B_R$, with $|D\eta |\le C/(R-r)$, yieldsAt $R=1$ a compactly supported cutoff can still be chosen with this bound by leaving an outer margin; alternatively pass to the limit from smaller radii.

For Moser iteration, set ${\beta }_j=p{\sigma }^j$ and $r_j=1/2+2^{-j-1}$. Repeated use of the preceding estimate bounds the successive norms byThis product is finite becauseAs the exponents tend to infinity, their norms on the Euclidean ball of radius $1/2$ tend to its essential supremum. ThereforeThe power $\beta$ is independent of the exponents of Hölder continuity in earlier questions.

For any $R>0$, set $U_R(y)=u(Ry)$ and $a_R^{ij}(y)=a^{ij}(Ry)$. These coefficients retain the same bounds and ellipticity, and a change of variables in the weak formulation shows that $U_R$ is a nonnegative weak subsolution on $B_1$. Part (a) gives the scale-correct estimateThe constant is independent of $R$. Fix a Euclidean ball $B_s$ and let $R\to \infty$ with $R>2s$. The right side tends to zero, so $u=0$ almost everywhere on $B_s$. Since $s$ is arbitrary, the Liouville theorem for integrable nonnegative elliptic subsolutions provesThis is $u\equiv 0$ as a Sobolev space element. Any continuous representative is identically zero pointwise; an arbitrary measurable representative can be modified at isolated points without changing the weak equation. Integrability and the explicit scaling force the conclusion.

Absolute values are missing in the printed formula. For signed real blocks, the intended block mixed norm must beWithout them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative scalar has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a norm and does not define the claimed general convex problem. Even integer exponents do not fix all other $q$ in the stated range.

With the corrected norm, let $q^{\prime }$ be the Holder conjugate exponent: $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Blockwise Holder inequality, with equality from a norming block, givesThe duality of block mixed norms identifies the closed compact convex set and the projection residual from the Moreau decompositionFor $q=2$, the dual blocks are Euclidean balls; for $q=1$, they are cubes given by componentwise bounds $|p_{i,j}^r|\le \lambda$. No injectivity or surjectivity of $A$ is required: compactness of the block product makes its linear image compact, and the strictly convex fidelity still gives a unique primal minimizer. For general $q^{\prime }$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.

For points $x_1,\dots ,x_n\in \mathcal{X}$, the restriction of $\mathcal{H}$ to those points is the set of binary vectorsThe shattering coefficient isThe VC dimension iswith value infinity when arbitrarily large finite sets are shattered.

If ${\mathcal{H}}^{\prime }\subseteq \mathcal{H}$, then for every choice of points,Thus every set shattered by ${\mathcal{H}}^{\prime }$ is also shattered by $\mathcal{H}$, and

Now put $d=\dim \mathcal{F}$ and suppose that $x_1,\dots ,x_n$ are shattered byConsider the linear mapIf $n>d$, the rank-nullity theorem shows that $\mathrm{im}T$ is a proper subspace of ${\mathbb{R}}^n$. Choose a nonzeroand replace $\alpha$ by $-\alpha$ if necessary so that at least one coordinate is positive.

Ask for the labeling that assigns label zero when ${\alpha }_i>0$ and label one when ${\alpha }_i<0$; coordinates with ${\alpha }_i=0$ may be labeled arbitrarily. Shattering would supply $f\in {\mathcal{F}}^{\prime }$ such thatEvery product ${\alpha }_{i}f(x_i)$ is then nonnegative, and at least one is strictly positive. Hencecontradicting $\alpha \perp \mathrm{im}T$. Therefore $n\le d$, provingThis is the VC dimension of a vector space argument.

Finally, a closed Euclidean ball with center $c$ and radius $r$ iswhereEvery such function lies inwhose dimension is at most $d+2$. Applying the result just proved gives the VC dimension upper bound for Euclidean balls

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

The indicators of closed Euclidean balls in ${\mathbb{R}}^d$ have VC dimension at most $d+2$. Indeed,wherebelongs to the span of $\Vert x{\Vert }_2^2,x_1,\dots ,x_d,1$.

A subset of the radius-$R$ Euclidean ball admits an $\epsilon$-metric net of size at most $(1+2R/\epsilon {)}^n$. Take a maximal separated set: its disjoint radius-$\epsilon /2$ Euclidean balls lie in the radius-$(R+\epsilon /2)$ ball. Comparing Lebesgue measures gives the estimate. In particular, the unit sphere admits a net of radius one with at most $3^n$ points.