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Independence gives $P(X=Y)={\sum }_r{p}_r{q}_r={\mathbb{E}}_p[2^{{\log }_2{q}_X}]$. The given exponential convexity inequality therefore yieldsTaking complementary probabilities provesThe strict positivity of the entries of $q$ makes all logarithms appearing in this finite expectation well defined.

Conditional entropy is $H(U\mid V)={\sum }_{v}P(V=v)H(U\mid V=v)$. Factor the joint probabilities as $P(U=u,V=v)=P(V=v)P(U=u\mid V=v)$ and expand their logarithms. This gives $H(X,Y)=H(Y)+H(X\mid Y)$. Applying the same identity within each conditional law given $Z=z$ and averaging gives $H(X,Y\mid Z)=H(X\mid Z)+H(Y\mid X,Z)$.

Put $E=1_{\{X\ne Y\}}$ and $\epsilon =P(E=1)$. Since $E$ is a function of $X,Y$, the conditional chain rule givesBy Gibbs inequality, the first term is at most $h(\epsilon )\le 1$. When $E=0$, $Y$ determines $X$. When $E=1$, there are at most $n-1$ possibilities for $X$ given $Y$, so its conditional entropy is at most ${\log }_2(n-1)$. ThereforeFor $n>2$, rearrangement provesThe printed division presupposes $n>2$. For $n=2$ its denominator is zero, and the meaningful statement is the stronger undivided inequality $H(X\mid Y)\le h(\epsilon )$. For $n=1$ the variables agree surely.

Use the convention $0{\log }_{2}0=0$. On the positive support of $p$, the given logarithm inequality givesThe logarithm is strictly concave, so its supporting line at one has equality only there. More explicitly, the given global bound forces $c=1/{\log }_{e}2$ by differentiation at one, and $x-1-{\log }_{e}x$ vanishes only at one. Equality above therefore requires $p_r=q_r$ on the positive support, and no mass of $q$ outside it. Thus equality holds exactly when $p=q$.

For the uniform distribution $q_r=1/n$, the divergence is $D(p,q)={\log }_{2}n-H(p)$. Its nonnegativity gives $\boxed{H(p)\le {\log }_{2}n}$, with equality precisely for the uniform law.

A scalar Tonelli existence theorem states: if $f(x,y,p)$ is continuous, convex in $p$, and has a uniform superlinear lower bound $f(x,y,p)\ge \Theta (|p|)-C$ with $\Theta (r)/r\to \infty$, and at least one fixed-endpoint admissible absolutely continuous function has finite energy, then the energy attains its infimum in the fixed-endpoint $W^{1,1}$ class. One may take $\Theta$ nonnegative and convex. The lower bound makes derivatives uniformly integrable; endpoint control then gives uniform compactness of a minimizing sequence, and convexity supplies the required lower semicontinuity.

For the specified functional, the integrand is continuous, nonnegative, and convex in the derivative, withThe affine function $u_0(x)=2x-1$ has the required endpoints and finite energy. Thus a global minimizer exists. In fact any finite-energy minimizer belongs to $W^{1,52}$, since its derivative's 52nd power is integrable.

It is unique. The admissible class is convex, and the integrand is strictly convex in $(y,p)$: $|p{|}^{52}$ and $|y{|}^{2011}$ are strictly convex, with the latter multiplied by $e^x>0$. If two distinct admissible finite-energy functions minimized the energy, their midpoint would have strictly smaller energy. Distinct continuous representatives differ on a set of positive measure, so the strict integral inequality really applies. This contradiction proves uniqueness.

The variational terminology refers to the size of a neighborhood, rather than weak versus strong convergence in a Banach space. The definitions in the two numbered children giveThe second implication holds because smallness in $W^{1,\infty }$ implies smallness in the uniform norm. Reverse implications need not hold.

For differentiable $f$ in its slope variable, define the Weierstrass excess functionThe strengthened Weierstrass condition means that some $\delta >0$ givesIt requires support at nearby heights and nearby reference slopes, while the trial slope $q$ is arbitrary. Requiring this only exactly on the curve is the unstrengthened necessary condition.

A standard Weierstrass sufficient condition assumes the usual $C^2$ smoothness of $f$, $u\in \mathcal{A}\cap C^1$, the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened condition above. Explicitly, for some $c>0$,for zero-endpoint variations. Then $u$ is a strong local minimizer. If the excess is strictly positive for $q\ne p$ throughout the tube, the strong minimum is strict. The second-variation hypothesis supplies a neighboring field of extremals. Its slope $p(x,y)$ gives the comparison identityfor uniformly close admissible $v$. This explains how the condition controls competitors with large derivatives.

For the supporting inequality, assume first that $F$ is convex and fix $z,z_0$. For $0<t\le 1$,Subtract $F(z_0)$, divide by $t$ and let $t\downarrow 0$. Differentiability givesConversely assume this inequality for every pair. Put $w=(1-t)z_0+tz$ and apply it with base point $w$ to both $z_0$ and $z$. Multiply those two inequalities by $1-t$ and $t$ and add. The gradient terms cancel, proving $F(w)\le (1-t)F(z_0)+tF(z)$. Thus the supporting inequality characterizes a differentiable convex function.

For the integral inequality let $\bar{z}=|\Omega {|}^{-1}{\int }_{\Omega }u$. Apply the supporting inequality at $\bar{z}$ to $u(x)$ and integrate. Its affine term has integral $DF(\bar{z})\cdot ({\int }_{\Omega }u-|\Omega |\bar{z})=0$, yieldingThis is Jensen's inequality. Since $u$ is integrable, the supporting affine lower bound has integrable negative part, so the right side is well-defined as an extended real integral. If its positive part is infinite, the inequality remains valid.

Use the direct method in the calculus of variations. The affine function $1-x$ is admissible with finite energy, and the energy is nonnegative since $2010$ is even. Thus $m=\inf I$ is finite. Choose a minimizing sequence $u_n$ with uniformly bounded energy. ThenThe fixed value $u_n(0)=1$ and the Cauchy-Schwarz inequality give $\Vert u_n{\Vert }_{\infty }\le 1+\Vert u_n^{\prime }{\Vert }_2$ and $|u_n(x)-u_n(y)|\le \Vert u_n^{\prime }{\Vert }_2|x-y{|}^{1/2}$. Hence the sequence is bounded in $W^{1,2}$ and uniformly bounded and equicontinuous.

By reflexivity and the Arzelà-Ascoli theorem, pass to a subsequence with $u_n\rightharpoonup u$ in $W^{1,2}$ and $u_n\to u$ uniformly. The two limits coincide; uniform convergence also preserves $u(0)=1$, $u(1)=0$. Since all functions stay in a common bounded interval, uniform convergence gives $u_n^{2010}\to u^{2010}$ uniformly, and therefore convergence of the potential integrals. The derivative term is lower semicontinuous for weak convergence, soAdmissibility gives $I(u)\ge m$. Thus