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An operator $G$ is $\beta$-cocoercive when $⟨Gz-Gw,z-w⟩\ge \beta \Vert Gz-Gw{\Vert }^2$ with $\beta >0$. This strengthens monotonicity and implies a Lipschitz constant $1/\beta$ by the Cauchy-Schwarz inequality. Gradient maps of smooth convex functions provide an important converse through the Baillon–Haddad theorem.

For a matrix with at least two columns of unit Euclidean norm, its mutual coherence is the largest magnitude of an inner product between distinct columns. It measures the strongest pairwise ambiguity of linear measurements. The Cauchy-Schwarz inequality gives $0\le \mu \le 1$. The order-two restricted isometry constant equals the mutual coherence, because every two-column Gram matrix has eigenvalues $1\pm |⟨a_i,a_j⟩|$.

The quotient of a horizontal strip by a horizontal translation. The model $(\mathbb{R}/\mathbb{Z})\times (0,M)$ has conformal modulus of an annulus $M$. Averaging horizontal loop lengths and applying Cauchy-Schwarz inequality proves that the core-loop extremal length equals $1/M$.

Let $U\subset {\mathbb{R}}^d$ be bounded, $A\in {\mathbb{R}}^d$ satisfy $\Vert A{\Vert }_2<2$, and $f\in L^2(U)$. The bilinear formon $H_0^1(U)$ is bounded. The Cauchy-Schwarz inequality and $2ab\le a^2+b^2$ giveso it is a coercive bilinear form. The Lax-Milgram theorem therefore gives a unique weak solution of $-\Delta v+v+A\cdot \nabla v=f$ with zero Dirichlet boundary condition.

For a regular scalar model with score function $S_{\theta }$ and Fisher information $I(\theta )={\mathbb{E}}_{\theta }{S}_{\theta }^2$, an estimator $T$ with mean $\psi (\theta )$ satisfiesIndeed, differentiating ${\mathbb{E}}_{\theta }T=\psi (\theta )$ gives ${\mathrm{Cov}}_{\theta }(T,S_{\theta })={\psi }^{\prime }(\theta )$, and the Cauchy-Schwarz inequality gives the bound.

For the cube $Q=(0,L{)}^n$ and $u\in H^1(Q)$,This follows from the pairwise-difference identityand a coordinate-by-coordinate path from $x$ to $y$. The Cauchy-Schwarz inequality bounds the squared sum of its $n$ increments by $n$ times the sum of their squares, while the one-dimensional fundamental theorem of calculus bounds each integrated increment by $L^{n+2}\Vert D_{i}u{\Vert }_2^2$. Division by $2|Q|=2L^n$ gives the stated constant.

For the complete $\ell$-uniform hypergraph $K_r^{\ell }$, with $r\ge \ell \ge 2$, the bound is $\pi (K_r^{\ell })\le 1-1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$. In the complementary covering formulation, every $r$-set containing an edge forces asymptotic edge density at least $1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$. It follows from clique-extension incidence inequalities and Cauchy-Schwarz inequality estimates on link degrees.

For $h\in L_0^2(P)$, truncate $h$ to $h_m=\max (-m,\min (h,m))$ and subtract $P{h}_m$. Dominated convergence gives $h_m\to h$ in L2 space, and the Cauchy-Schwarz inequality gives $P{h}_m\to 0$. Thus bounded centered directions are dense in the mean-zero L2 space. The measure defining the L2 norm is essential: density-weighted L2 space can contain functions outside the unweighted Lebesgue space.

Write $e^u\varphi (u)=\int \psi (t)e^{-iut}dt$. The double integral $\iint \psi (t)\psi (t^{\prime })(1+it)(1+i{t}^{\prime })/(2+i(t+t^{\prime }))dtd{t}^{\prime }$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-{\varphi }^{\prime }(u)$, then use Fubini's theorem. For a real cutoff the resulting square is positive even though the Fourier transform factors have no complex conjugation. The Cauchy-Schwarz inequality and $\varphi (0)=1$, $\varphi (1/3)=0$ give $c_{\varphi }\ge 3$.

If $a\ne b$ and $x\cdot a=A$, $x\cdot b=B$, the Cauchy-Schwarz inequality applied to $x\cdot (a-b)=A-B$ gives $\Vert x\Vert \ge |A-B|/\Vert a-b\Vert$. Equality requires $x$ parallel to $a-b$ and satisfying both original equations. The estimate gives a lower bound on the distance from the origin to the intersection; it need not be sharp if that equality condition is incompatible with the constraints.

Let $X$ be a continuous martingale on $[0,1]$ with $X_0=0$ and $|X_t|\le C$. For the dyadic squared-increment sums, including the unfinished increment at time $t$, set $A_t^{(n)}={\sum }_k(X_{k{2}^{-n}\wedge t}-X_{(k-1)2^{-n}\wedge t}{)}^2$. The associated martingale transform $M_t^{(n)}=(X_t^2-A_t^{(n)})/2$ satisfies $\mathbb{E}(M_1^{(n)}{)}^2\le C^4$ and $\mathbb{E}(A_1^{(n)}{)}^2\le 10C^4$. Discrete martingale orthogonality, the path modulus of continuity and the Cauchy-Schwarz inequality show that the terminal transforms are Cauchy in $L^2$. The Doob L2 maximal inequality then gives $\mathbb{E}{\sup }_{t\le 1}|A_t^{(n)}-A_t^{(m)}{|}^2\to 0$. The sums need not be increasing in time before taking the limit. This provides an elementary construction of quadratic variation under boundedness.

Equality in the three-dimensional Loomis--Whitney inequality forces a measurable body to agree up to a null set with a Cartesian product $E_1\times E_2\times E_3$. This follows from the equality conditions in the two Cauchy-Schwarz inequalities used in its proof. If the body is connected and is a finite union of positive-volume axis-parallel boxes, each $E_i$ is an interval and equality up to a null set upgrades to exact equality with one box.

For a set partition $\mathcal{P}$ of the vertices of a graph of order $n$, definewhere the sum is ordered, the adjacency indicator defines the density also on diagonal pairs, and exceptional vertices are treated as singleton parts. Refinement cannot decrease this energy, by the Cauchy-Schwarz inequality. An irregular pair of equal cells of size $m$, with witnesses of relative size at least $\epsilon$ and density discrepancy greater than $\epsilon$, increases the energy by more than ${\epsilon }^4{m}^2/n^2$ for each orientation. Splitting refined atoms into equal chunks and leftover singleton parts is a further refinement, so restoring equitability causes no energy loss.

The Euclidean norm is self-dual:The upper bound is Cauchy-Schwarz inequality, and equality holds at $v=x/\Vert x{\Vert }_2$ when $x\ne 0$.

If ${\theta }_r$ have circular spacing at least $\delta$, thenMultiply the exponential sum by $e(-Mt)$, apply the Sobolev–Gallagher inequality on disjoint arcs of length $\delta$, and sum. The finite-interval Parseval identities and Cauchy-Schwarz inequality bound the derivative contribution by $2\pi N\sum |a_n{|}^2$.

On a compact interval, squared increments of a continuous finite variation process are bounded by its modulus of continuity at the mesh size times its total variation, and therefore vanish. The Cauchy-Schwarz inequality bounds the mixed increment sum by the square root of the product of the two squared-increment sums. The local-martingale sum is bounded in probability, so the mixed term also vanishes.

If $A(t)=A+tE$ and $\lambda (t)$ is a differentiable function defining a simple eigenvalue, thenThe Cauchy-Schwarz inequality and the operator norm give

Let $I=[n]$ for odd $n\ge 3$, embed it in the cyclic group $G=\mathbb{Z}/(2n+1)\mathbb{Z}$, and let $A\subseteq I$ have density of a finite subset $\alpha$. If $A$ has no nonconstant three-term arithmetic progression and $n\ge 4{\alpha }^{-2}$, its balanced indicator function $f=1_A-\alpha 1_I$ satisfiesIndeed, the normalized trilinear arithmetic progression count has the Fourier analysis on a finite abelian group formula $\Lambda (h_1,h_2,h_3)={\sum }_r{\hat{h}}_1(r){\hat{h}}_2(-2r){\hat{h}}_3(r)$. Its values on $1_A$ and $\alpha 1_I$ are $\alpha n/|G{|}^2$ and ${\alpha }^3(n^2+1)/(2|G{|}^2)$. Telescoping their difference into three terms containing $f$, the Parseval identity and Cauchy-Schwarz inequality bound its magnitude by $3\max |\hat{f}|\alpha n/|G|$. The difference is at least ${\alpha }^3{n}^2/(4|G{|}^2)$, giving the claim. The zero Fourier coefficient vanishes by the definition of $\alpha$.

In the normal linear model, assume $\epsilon \sim N_n(0,{\sigma }^{2}I)$ is independent of $X$. Independent rows of $X$ with multivariate normal distribution $N_p(0,I_p)$ give a simple random design for Debiased Lasso. Let $p\to \infty$, $\log p=o(n)$ and $s\log p=o(\sqrt{n})$, with fixed positive noise scale. Choose the Lasso penalty and all Nodewise Lasso penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with probability tending to one. Column squared norms divided by $n$ converge uniformly to one. To check the Compatibility condition for the Lasso directly, put $\hat{\Sigma }=X^{T}X/n$ and $\eta ={\max }_{j,k}|{\hat{\Sigma }}_{jk}-1_{\{j=k\}}|$. On the Lasso cone condition, $\Vert \delta {\Vert }_1\le 4\Vert {\delta }_S{\Vert }_1$, and the Cauchy-Schwarz inequality givesFor $s\ge 1$, entrywise concentration of a Gaussian Gram matrix gives $\eta \le D_0\sqrt{\log p/n}$ with probability tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta \to 0$ on this event, so one can take ${\varphi }^2\ge 1/2$ eventually. The conditional Gaussian tail bound for the noise score and the Basic inequality for the Lasso then give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le Cs\sqrt{\log p/n}$ with probability tending to one. Consequently the debiased-Lasso remainder bound is at most $As\log p/\sqrt{n}$ with probability tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.

The Cauchy-Schwarz inequality gives