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Let the square contain the Euclidean ball of radius $R$, set $r_x=\max (1,\Vert x\Vert )$, and define the logarithmic cutoffThen ${\phi }_0=1$ and $\phi =0$ on the boundary. On an edge at radius comparable to $r$, the mean value theorem gives $|{\phi }_x-{\phi }_y|\le C/(r\log R)$. There are $O(r)$ edges in the annulus of radius $r$, hence the discrete Dirichlet energy satisfiesThis logarithmic cutoff is the discrete manifestation of recurrence in two dimensions.

The Liouville approximation theorem states that if $\alpha$ is an irrational algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such that every reduced rational $p/q$ satisfies

Let $P\in \mathbb{Z}[X]$ be the minimal polynomial of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne 0$. Moreover,and henceOn the compact interval $I=[\alpha -1,\alpha +1]$, putIf $p/q\in I$, the mean value theorem gives a point $\xi$ between $\alpha$ and $p/q$ such thatThus $|\alpha -p/q|\ge M^{-1}{q}^{-d}$. If $p/q\notin I$, then $|\alpha -p/q|>1\ge q^{-d}$. Taking anyproves the stated strict inequality.

Since $U\subset {\mathbb{R}}^3$ is bounded and smooth, the Sobolev embedding theorem gives continuous embeddings $H^1(U)\hookrightarrow L^4(U)$ and $H^1(U)\hookrightarrow L^6(U)$. Because $|\sin s|\le 1$,Taking the $L^2$ norm in time gives

For $F(s)=s^2\sin s$, the mean value theorem and $|F^{\prime }(s)|\le 2|s|+s^2$ implyApply the Generalized Holder inequality in space, using $L^4$ for the quadratic products and $L^6\cdot L^3$ for the cubic products, and then use the two Sobolev embeddings. Pointwise in time this yieldsTaking the $L^2$ norm in time proves the required estimate with the factor $\gamma T^{1/2}$.

If $f\prime \ge 0$, the mean value theorem gives $f(y)-f(x)=f\prime (c)(y-x)\ge 0$. Conversely, if $f$ is increasing, every difference quotient with positive or negative increment is nonnegative; taking its limit gives $f\prime (x)\ge 0$.

For $x>0$, apply the mean value theorem to $[0,x]$: $f(x)/x=f\prime (c)$ for some $c<x$. Since $f\prime$ is increasing, $f(x)/x\le f\prime (x)$. Thereforeand part (c) proves that the quotient is increasing.

Put $\Lambda =n\log 2-m\log 3$. It is nonzero by unique prime factorization. If $3^m\le 2^{n-1}$ or $3^m\ge 2^{n+1}$, the claimed inequality follows immediately after increasing the effective constant. We may therefore assumewhich implies $m\asymp n$.

The Baker lower bound for a homogeneous linear form in logarithms, with the fixed algebraic numbers $2$ and $3$, givesfor an effective absolute constant $C_0$. If $|\Lambda |>1$, the desired conclusion is again immediate. Otherwise, the mean value theorem applied to the exponential function on $[-1,1]$ gives $|e^u-1|\ge c|u|$. HenceAbsorbing the fixed factor into a larger exponent proves $|2^n-3^m|\ge 2^n/n^C$.

The mean value theorem gives $f(b)-f(a)=(b-a)f\prime (c)>0$.

If $\alpha >1/2$, monotonicity already gives $\mathbb{E}{f}^{(1/2)}\ge \alpha >1/2$, so assume $\alpha \le 1/2$. Suppose for a contradiction that $\mathbb{E}{f}^{(1/2)}\le 1/2$. By the mean value theorem, some $s\in (p,1/2)$ satisfiesThe Margulis–Russo formula identifies this derivative with the appropriately normalized total influence, so ${\mathbf{I}}_s(f)$ is bounded solely in terms of $\zeta$. The $p$-biased Friedgut junta theorem then supplies, for any small $\eta >0$, a Boolean $J$-junta $h$ with $|J|\le r(\zeta ,\eta )$ and

Because $f$ is monotone, ${\mathbb{P}}_s(f=1)\le 1/2$, hence ${\mathbb{P}}_s(h=0)\ge 1/2-\eta \ge 1/4$ when $\eta \le 1/4$. It follows thatFor some assignment $u$ on $J$ with $h(u)=0$, therefore, ${\mathbb{E}}_s{f}_u\le 4\eta$. Monotonicity and $p<s$ imply ${\mathbb{E}}_p{f}_u\le 4\eta$. Choose $\eta <\alpha /5$, set $\epsilon =\eta$, and take $r\ge |J|$. Thencontradicting $(\epsilon ,p,r)$-quasirandomness. Thus $\mathbb{E}{f}^{(1/2)}>1/2$.

Choose a radial mollifier $\rho \in C_c^{\infty }$ with integral one and support in the unit ball, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one there. For a radial tempered distribution $T$, the functionsare radial Schwartz functions. Smoothness comes from convolution of a tempered distribution with a Schwartz function, and compact support comes from the cutoff. Pairing with $\phi$ gives $⟨f_j,\phi ⟩=⟨T,{\check{\rho }}_{1/j}\ast (\chi (\cdot /j)\phi )⟩$. With $q_m(\phi )={\max }_{|\beta |\le m}{\sup }_x(1+|x|{)}^m|{\partial }^{\beta }\phi (x)|$, the cutoff tail and the mean value theorem giveThe continuity estimate for $T$ therefore proves $f_j\to T$ even in the strong dual topology. Inserting a cutoff is essential because convolution alone need not give rapid decay.

For positive rates $a,b$, the mean value theorem in the rate gives a bound by $|a-b|x{e}^{-\min (a,b)x}$. Its maximum is $|a-b|/[e\min (a,b)]$. Combining this estimate with coefficient errors bounds a finite spectral sum uniformly on a half-line, including slow scales much longer than one. Positivity of both rates is essential: unstable exponentials need a different time-window or relative-error analysis.

If a twice differentiable real function on $\mathbb{R}$ satisfies $f^{\prime \prime }(x)\le -c$ for some fixed $c>0$, then it has a unique global maximum. The mean value theorem shows that $f^{\prime }$ is strictly decreasing, and that $f^{\prime }(x)+cx$ is nonincreasing. Hence $f^{\prime }(x)\le f^{\prime }(0)-cx$ for $x>0$ and $f^{\prime }(x)\ge f^{\prime }(0)-cx$ for $x<0$. These inequalities force opposite derivative signs at the two ends. The intermediate value theorem supplies a unique zero of $f^{\prime }$, at which $f$ changes from increasing to decreasing. Merely having $f^{\prime \prime }<0$ does not guarantee existence: $f(x)=-e^{-x}$ is increasing and approaches its unattained supremum zero.