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For a periodic absolutely continuous function with derivative in $L^2$, integration by parts gives $\widehat{f^{\prime }}(n)=in\hat{f}(n)$. Bessel's inequality and the Cauchy-Schwarz inequality prove the displayed bound in the normalized circle norm. Adding the constant coefficient gives an absolutely and uniformly convergent Fourier series. Merely bounding individual coefficients by $O(1/|n|)$ would not establish absolute convergence.

For a degree-$N$ rational map, the angular Jacobian of a rational map has average $N$, so the Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$. In dimensionless Skyrme model units the radial energy within the rational-map ansatz isFirst minimize $\mathcal{I}$ over admissible maps, then minimize over profiles with $f(0)=\pi$, $f(\infty )=0$. This gives a restricted variational approximation and an upper bound on the unrestricted minimum in that topological sector, not an exact multi-Skyrmion solution. Houghton, Manton and Sutcliffe developed this construction in arxiv.org/abs/hep-th/9705151.

Let $\alpha$ be a badly approximable number, and sum $B={\sum }_{D<d\le 2D}{\sum }_{W<w\le 2W,dw\le X}{a}_d{b}_{w}e(\alpha dw)$ with $|a_d|\le \tau (d)$ and $|b_w|\le 1$. The Cauchy-Schwarz inequality, divisor-square summatory bound and geometric sums give $|B{|}^2\ll D{\log }^{3}X(DW+W{\sum }_{h\le W}\min (D,\Vert h\alpha {\Vert }^{-1}))$. Separated rotations bound the last sum by $O_{\alpha }(W\log (2W))$, giving the displayed estimate. A hyperbolic cutoff remains valid because the allowed first-variable values after expansion form an interval.

With uniform expectations and real-valued functions,Apply the Cauchy-Schwarz inequality first in $y$. Expand the remaining square, and apply the Cauchy-Schwarz inequality in $(x,x^{\prime })$ to $u(x)u(x^{\prime })$ and ${\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. Their squared averages are $\Vert u{\Vert }_2^4$ and $\Vert f{\Vert }_{\square }^4$, respectively.

Two applications of the Cauchy-Schwarz inequality giveThe fourth moment exposes differences of squares that can be factored and estimated by one-dimensional geometric sums.

If a bipartite graph with parts $X,Y$ has at least $\delta |X||Y|$ edges, then the number of ordered tuples $(x_1,x_2,y_1,y_2)\in X^2\times Y^2$ for which every $x_i{y}_j$ is an edge is at leastIndeed, if $d(x_1,x_2)$ is the number of common neighbours of $x_1,x_2$ in $Y$, two applications of the Cauchy-Schwarz inequality give

For every $\epsilon >0$ there is a positive integer $K(\epsilon )$ such that every finite bipartite graph has an $\epsilon$-regular pair of partitions with at most $K(\epsilon )$ cells on each side.

To prove this, give $X\times Y$ the uniform probability measure, let $f$ be the indicator function of the edge set, and for partitions $\mathcal{P},\mathcal{Q}$ define their energy byThis conditional expectation is an orthogonal projection, so $0\le \mathcal{E}\le \Vert f{\Vert }_2^2\le 1$. If the partitions are not $\epsilon$-regular, choose witnesses $A_{ij}\subseteq X_i$ and $B_{ij}\subseteq Y_j$ in every irregular cell pair and refine each $X_i$ by all its $A_{ij}$ and each $Y_j$ by all its $B_{ij}$. The refined energy exceeds the old energy byOn $A_{ij}\times B_{ij}$ the absolute value function of the mean of the difference is greater than $\epsilon$, while this rectangle occupies at least an ${\epsilon }^2$ fraction of $X_i\times Y_j$. The Cauchy-Schwarz inequality therefore contributes more than ${\epsilon }^4|X_i||Y_j|/(|X||Y|)$ from that cell pair. Irregular pairs have total weight greater than $\epsilon$, so every refinement raises the energy by more than ${\epsilon }^5$. The process stops after at most $\lceil {\epsilon }^{-5}\rceil$ refinements. If the current cell counts are $r,s$, the construction gives at most $r{2}^s,s{2}^r$ new cells, so a finite iterated bound depending only on $\epsilon$ supplies $K(\epsilon )$.

For four real-valued functions on a finite Cartesian product, let $\Lambda (f_{00},f_{01},f_{10},f_{11})={\mathbb{E}}_{x_0,x_1,y_0,y_1}{\prod }_{i,j=0}^1{f}_{ij}(x_i,y_j)$. Repeated Cauchy-Schwarz inequalities giveFirst separate the two $y$ averages and apply Cauchy-Schwarz inequality in $(x_0,x_1)$. Each resulting squared factor is ${\mathbb{E}}_{y,y^{\prime }}({\mathbb{E}}_{x}f(x,y)f(x,y^{\prime }))({\mathbb{E}}_{x}g(x,y)g(x,y^{\prime }))$, bounded by $\Vert f{\Vert }_{\square }^2\Vert g{\Vert }_{\square }^2$ by another Cauchy-Schwarz inequality. Expanding the four factors of $f+g$ and applying this inequality to each of the sixteen terms gives the triangle inequality for the box norm.

For a complete orthonormal basis of $L^2({\mathbb{R}}_{+})$, the Parseval identity for a Hilbertian basis identifies the sum of the feature products with $⟨1_{[0,s]},1_{[0,t]}⟩=\min (s,t)$. The Cauchy-Schwarz inequality gives absolute convergence.

For total exposure $W={\sum }_j{m}_j$ and weighted average ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$, the best affine mean squared error estimate of the conditional mean is $Z{\overline{X}}_w+(1-Z)m_0$, where $Z=Wa/(Wa+v)$. If the coefficients sum to $B$, their error is $a(1-B{)}^2+v{\sum }_j{b}_j^2/m_j$. Cauchy-Schwarz inequality minimizes the latter term at $b_j=B{m}_j/W$, and a one-dimensional quadratic minimization then gives $B=Z$.

Let $\mathcal{G}$ be the first Gaussian chaos of a centered Gaussian random variable in a Banach space. By Fernique's theorem, $Sg=\mathbb{E}[Xg]$ is a Bochner integral in $B$. The map $S$ is injective, because $Sg=0$ implies $\mathbb{E}[g\ell (X)]=0$ for every continuous linear functional $\ell$, hence $g=0$. Its range becomes a Hilbert space under $⟨Sg,Sv{⟩}_H=\mathbb{E}[gv]$. The embedding in $B$ is continuous by Cauchy-Schwarz inequality. Its reproducing property is $⟨h,S\ell (X){⟩}_H=\ell (h)$. Thus this is the Reproducing-kernel Hilbert space of the random variable, including degenerate laws. For $h=Sg$, the scalar coordinate $\hat{h}(X)=g$ has variance $\Vert h{\Vert }_H^2$; it is generally a measurable linear coordinate rather than a continuous functional on $B$.

The canonical height pairing is $⟨P,Q⟩=(\hat{h}(P+Q)-\hat{h}(P)-\hat{h}(Q))/2$. The height parallelogram identity makes it bilinear. It descends through torsion and is positive definite on the free part of the Mordell-Weil group. The Cauchy-Schwarz inequality gives $\hat{h}(P-Q)\le 2\hat{h}(P)+2\hat{h}(Q)$, useful for height descent.

For an elliptic curve over a number field, its nonnegative canonical height of an elliptic curve has finite bounded subsets by the Northcott theorem and its bounded difference from half the naive logarithmic height. By the Weak Mordell-Weil theorem, choose representatives $R_i$ modulo $2E(K)$ and put $H={\max }_i\hat{h}(R_i)$. If $P=2Q+R_i$, the Cauchy-Schwarz inequality for the height pairing gives $\hat{h}(Q)\le (\hat{h}(P)+H)/2$. Iterating brings every point into the finite set of height at most $H+1$. This set and the $R_i$ generate $E(K)$, proving the Mordell-Weil theorem without assuming finite generation when constructing the height.

Choose a directed path ${\eta }_{xy}$ of positive-probability transitions for each ordered pair of states in a finite reversible Markov chain. Write $Q(e)=\pi (u)P(u,v)$ for a directed edge $e=(u,v)$, and let $N_e(\eta )$ count its occurrences. DefineThe Cauchy-Schwarz inequality bounds $(f(x)-f(y){)}^2$ by $|{\eta }_{xy}|$ times the sum of squared differences along the path. Multiply by $\pi (x)\pi (y)/2$, sum and interchange finite sums. The identity ${\mathrm{Var}}_{\pi }(f)=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves the bound. This convention counts directed edges and retains the factor $1/2$ in the Dirichlet form of a Markov chain.

For a finite simple graph, $\alpha (G)\ge {\sum }_{v\in V(G)}1/(1+\mathrm{deg}v)\ge |V(G){|}^2/(|V(G)|+2|E(G)|)$. Randomly order its vertices and retain each vertex preceding all its graph neighbours. This gives an independent set with the first expression as its expected value. The second inequality is the Cauchy-Schwarz inequality.

If $r=dQ/dP$, independence gives $E_{P^{\otimes n}}{\prod }_{i}r(X_i{)}^2=(E_P{r}^2{)}^n$. Since $E_{P}r=1$, this proves the displayed identity. The Cauchy-Schwarz inequality also gives $\Vert Q^{\otimes n}-P^{\otimes n}{\Vert }_1^2\le {\chi }^2(Q^{\otimes n}\Vert P^{\otimes n})$. Thus perturbations of size ${\chi }^2(Q\Vert P)=O(1/n)$ have a uniformly bounded joint total variation distance, a useful input to a metric squared-loss two-point bound.

The Cauchy-Schwarz inequality gives $\Vert P-Q{\Vert }_{\mathrm{TV}}=\frac{1}{2}{\mathbb{E}}_Q|dP/dQ-1|\le \frac{1}{2}\sqrt{{\chi }^2(P\Vert Q)}$. Every statistical hypothesis testing rule has sum of its Type I error and Type II error at least $1-\Vert P-Q{\Vert }_{\mathrm{TV}}$, proving the displayed bound. A mixture model of alternatives yields a lower bound on the worst alternative error because a maximum dominates an average.