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The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

With $\Sigma =I_d$, constrained likelihood maximization minimizes $\Vert \theta -\overline{X}{\Vert }^2$. On the unit sphere this is $1+\Vert \overline{X}{\Vert }^2-2{\theta }^T\overline{X}$. By the Cauchy-Schwarz inequality its minimizer isIf $\overline{X}=0$, every unit vector is a maximum-likelihood estimator; a specified measurable tie-breaking choice makes an estimator defined everywhere. This event has probability zero in the nonsingular normal model, but the optimization still has that exceptional case.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

Using $\max \{X^2,Y^2\}=(X^2+Y^2+|X^2-Y^2|)/2$ and Cauchy-Schwarz inequality,

The Cauchy-Schwarz inequality is $|\mathbb{E}(XY){|}^2\le \mathbb{E}(X^2)\mathbb{E}(Y^2)$. Markov inequality says that for $X\ge 0$ and $a>0$, $\mathbb{P}(X\ge a)\le \mathbb{E}X/a$.

Jensen inequality states that for convex $\varphi$,Let $\mu =\mathbb{E}X$. A supporting line at $\mu$ gives $\varphi (x)\ge \varphi (\mu )+c(x-\mu )$. Taking expectations makes the linear term vanish.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

Put $I={\int }_0^{1}f(t)dt$. If $I\ne 0$, take the inner product with the unit vector $u=I/\Vert I{\Vert }_2$. Thenby the Cauchy-Schwarz inequality; the case $I=0$ is immediate. This is the triangle inequality for a vector-valued integral.

Equality in the Euclidean argument requires $u\cdot f(t)=\Vert f(t){\Vert }_2$ for every $t$: the nonnegative continuous difference has integral zero. Equality in Cauchy-Schwarz inequality then says $f(t)=\lambda (t)u$ with $\lambda (t)\ge 0$. Conversely, if $f(t)=\lambda (t)v$ for one fixed $v\in {\mathbb{R}}^m$ and a continuous $\lambda \ge 0$, then for every norm,Thus these nonnegative-ray functions, including $f\equiv 0$, are exactly the required functions.

The Lax-Milgram theorem assumes that $H$ is a real Hilbert space and $B:H\times H\to \mathbb{R}$ is a bounded bilinear form and a coercive bilinear form: there are $M,\alpha >0$ such thatThen for every continuous linear functional $\ell \in H^{\prime }$ there is a unique $u\in H$ satisfyingSymmetry of $B$ is not required.

For the proof, the Riesz representation theorem provides a bounded linear operator $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v{)}_H$ and $\ell (v)=(g,v{)}_H$. Coercivity and the Cauchy-Schwarz inequality implyThus $T$ is injective and its range is closed: if $T{u}_j$ converges, the displayed inequality makes $u_j$ a Cauchy sequence; completeness gives $u_j\to u$ and $T{u}_j\to Tu$. If $w$ lies in the orthogonal complement of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha \Vert w{\Vert }^2\le B[w,w]=0$, so $w=0$. A closed subspace with zero orthogonal complement is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives

For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

Statement and definitions. For disjoint nonempty sets $A,B$ of vertices, their edge density of a bipartite graph is $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair of vertex sets, or an $\epsilon$-uniform pair, ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$ and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph of order $n\ge n_0$ has a set partitionwith $m_0\le k\le M$, $|V_0|\le \epsilon n$, equal nonzero sizes $|V_1|=\cdots =|V_k|$, and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $1\le i<j\le k$, that fail to be $\epsilon$-uniform. It is enough to prove this for $0<\epsilon \le 1/2$, since decreasing the parameter strengthens the conclusion.

Energy increment. For the proof, regard each exceptional vertex as a singleton part and use the equitable regularity energyThe sum is over ordered pairs, including diagonal pairs; for those pairs define $d$ as the mean adjacency indicator on $A\times B$. Thus $0\le q\le 1$. Under a refinement, the subrectangle densities have the original density as their weighted mean. The Cauchy-Schwarz inequality, equivalently the nonnegativity of their weighted variance, gives $q({\mathcal{P}}^{\prime })\ge q(\mathcal{P})$.

Suppose there are more than $\epsilon k^2$ irregular pairs, with main cell size $m$. For each such pair choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$ and with $|d(X,Y)-d(V_i,V_j)|>\epsilon$. Refine every main cell by all its witness subsets. Each splits into at most $2^{k-1}$ atoms, and we may use the larger bound $2^k$.

For an ordered irregular rectangle, write its refined densities as $d_{ab}$ and relative areas as $w_{ab}$. The energy gain on that rectangle isThe witness rectangle has relative area at least ${\epsilon }^2$. Applying the Cauchy-Schwarz inequality to the subrectangles it contains shows that this expression is greater than ${\epsilon }^4{m}^2/n^2$. There are two orientations of each irregular pair. Since $km=n-|V_0|\ge n/2$, the total gain is greater than

Restoring equal sizes without losing energy. Put $\ell =\lfloor m/4^k\rfloor$, assuming $m\ge 2\cdot 4^k$. Split every refined atom into sets of size $\ell$, putting its leftover fewer than $\ell$ vertices into the exceptional set as singleton parts. This is a further refinement, so the energy cannot decrease. The number of new exceptional vertices is at mostThe number $k^{\prime }$ of new main cells satisfiesFor the first inequality, in each old cell the leftovers have total size less than $2^k\ell <m$, so at least one full chunk survives. All previous exceptional singleton parts are kept. Treating leftovers as singleton parts is what makes the energy increment survive equitabilization exactly.

Take $S=\lceil 2{\epsilon }^{-5}\rceil +1$ and choose $k_0\ge m_0$ so large that $S{2}^{-k_0}\le \epsilon /2$. Start with $k_0$ equal main cells, with fewer than $k_0$ exceptional vertices. Define a finite bound $M$ by iterating $x\mapsto 2x{4}^x$ a total of $S$ times, starting from $k_0$. Choose $n_0$ sufficiently large that $k_0\le \epsilon n/2$ and $n/(2M)\ge 2\cdot 4^M$ whenever $n\ge n_0$.

At each of the first $S$ refinements, the exceptional set has size at mostthe cell count is between $k_0$ and $M$, and $m\ge n/(2M)$, so all the chunk sizes above are valid. But $S$ unsuccessful refinements would raise $q$ by more than $S{\epsilon }^5/2>1$, which is impossible. Thus the procedure terminates at a partition satisfying the Szemerédi regularity lemma.

Three linearly large uniform sets. Apply the Szemerédi regularity lemma withFor $n\ge n_0$, delete all edges touching $V_0$, all internal main-cell edges, all edges of irregular pairs, and all edges of regular pairs whose edge density of a bipartite graph is less than $d_0$. The respective losses are bounded byThus the total loss is at most $0.00102n^2$, and the remaining graph has at leastedges. Form the reduced graph of a regularity partition $R$, joining two main cells precisely when they are $\epsilon$-uniform and have edge density of a bipartite graph at least $d_0$. If their common size is $m$, the remaining edge count is at most $e(R)m^2$. Since $km\le n$, we obtain $e(R)>k^2/4$. The Mantel theorem gives a triangle in a graph in $R$. Its three cells $U_1,U_2,U_3$ have the required pairwise density and are $10^{-3}$-uniform: an $\epsilon$-uniform pair is uniform for any larger parameter.

Each selected cell has size $m\ge (1-\epsilon )n/M$. ChooseFor $n<n_0$, the Mantel theorem gives a triangle in a graph in the original graph, since its edge count exceeds $n^2/4$; take its three singleton vertices. Their pairs have density one and are uniform for every positive parameter, and $1>cn$. Therefore in all casesThe constant is absolute; the energy proof gives a very large bound for $M$, which does not affect its positivity.

Use the energy of a curve normalizationFor a variation $F(s,t)$ put $T={\partial }_{t}F$ and $V={\partial }_{s}F$. The Levi-Civita connection is torsion free, so ${\nabla }_{s}T={\nabla }_{t}V$. Differentiating energy once gives $E^{\prime }=[⟨V,T⟩{]}_a^b-\int ⟨V,{\nabla }_{t}T⟩$. Along a geodesic, ${\nabla }_{t}T=0$. Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, consistent with the specified positive sectional curvature. Indeed ${\nabla }_s{\nabla }_{t}V={\nabla }_t{\nabla }_{s}V+R(V,T)V$, and $⟨R(V,T)V,T⟩=-⟨R(V,T)T,V⟩$. The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form $I(V,V)$.

Let $\dim M=2m$. Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent $T$, and its restriction to $T^{\perp }$ is an orientation-preserving orthogonal map of odd dimension $2m-1$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue $+1$. Choose a nonzero fixed vector $v$ normal to $T$ and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field $V$ with ${\nabla }_{t}V=0$.

For the exponential variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$, the endpoints match periodically. Strictly positive sectional curvature givesThus $E({\gamma }_s)<E(\gamma )$ for small nonzero $s$. By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.

For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.

Set $M_n={\sum }_{j=0}^n{X}_j$ and ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. The independence and zero expected values give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$, so $(M_n)$ is a martingale. The variance additivity for independent random variables givesThe Cauchy-Schwarz inequality implies ${\sup }_n\mathbb{E}|M_n|\le \sqrt{V}$. Applying the Martingale convergence theorem from (b),In fact the L2 martingale convergence theorem also gives convergence in L2. Independently, for $n>m$ the same variance additivity for independent random variables yields $\mathbb{E}|M_n-M_m{|}^2={\sum }_{j=m+1}^n\mathbb{E}[X_j^2]\to 0$, confirming that the partial sums form a Cauchy sequence in $L^2$.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

For a diagonal entry, put $Z_j={\sum }_{i=1}^d(A_{ij}^2-1)$. Independence and the given moment-generating function inequality for the chi-squared distribution giveThe Chernoff bound with $a=t/4$ yieldsApply the same argument to $-Z_j$ and sum the two tails:

To obtain the off-diagonal bound, set $U=(V+W)/\sqrt{2}$ and $D=(V-W)/\sqrt{2}$. They are jointly normal with unit variances and covariance zero, so independence of uncorrelated jointly normal variables makes them independent standard normals. Since $VW=(U^2-D^2)/2$, the moment-generating function of the product of two independent standard normal random variables isFor $j\ne k$, $(A^{T}A{)}_{jk}$ is a sum of $d$ independent such products. The given inequality implies $\mathbb{E}{e}^{a(A^{T}A{)}_{jk}}\le e^{d{a}^2}$ for $|a|\le 1/2$. Taking $a=t/2$, and repeating for the negative tail, gives

Now take $t=c\sqrt{\log p/d}\in (0,1)$ and $E=A^{T}A/d-I_p$. There are $p$ diagonal entries but only $p(p-1)/2$ distinct off-diagonal entries, since $E$ is symmetric. By the union bound, the event $\Vert E{\Vert }_{\max }\le t$ has probability at leastThis is the required entrywise concentration of a Gaussian Gram matrix.

On that single event, any vector $z$ supported on at most $s$ coordinates satisfieswhere the last step is the Cauchy-Schwarz inequality on the support. If $u,v$ each have at most $s/2$ nonzero coordinates, $z=u-v$ has at most $s$. The same entrywise event applies simultaneously to all of them, so no union over vectors is needed. The sparse norm preservation from entrywise Gram control givesThe printed ratio is undefined at $u=v$. Its equivalent quadratic statement, $|\Vert A(u-v){\Vert }_2^2/d-\Vert u-v{\Vert }_2^2|\le st\Vert u-v{\Vert }_2^2$, includes that case trivially. The nontrivial concentration argument assumes $p\ge 2,c>0$; at $p=1$ the stated probability lower bound is nonpositive and hence vacuous.

Let $d=d_G(a,b)>0$, and for $i=1,\dots ,d$ putEach edge cutset ${\Pi }_i$ separates $a$ from $b$. Adjacent vertices have graph distances from $a$ differing by at most one. Hence an edge can cross at most one of these level boundaries, so the edge cutsets are pairwise edge-disjoint. Write $k_i=|{\Pi }_i|\ge 1$; then ${\sum }_i{k}_i\le |E|$.

The Nash-Williams inequality for unit conductances giveswhere the middle inequality is the Cauchy-Schwarz inequality. One can also derive the first inequality directly: every unit flow has signed net flux one through each ${\Pi }_i$, so its energy of a flow on that cut is at least $1/k_i$; sum over the disjoint cuts and use the Thomson principle.

Combining with the commute time identity yieldsThe case $a=b$ is trivial. A path graph, with $a,b$ its endpoints, attains equality.