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For a single-phase ideal gas mixture with molar amounts $N_i$, common temperature $T$, and pressure $P$, extensivity and ideal mixing giveHere $N={\sum }_i{N}_i$, $x_i$ is the mole fraction, $\mathcal{R}$ is the molar gas constant, and $P^{\circ }$ is the standard pressure. Integrating $\partial {\mu }_i/\partial P=\mathcal{R}T/P$ and including the ideal mixing entropy proves ${\mu }_i={\mu }_i^{\circ }+\mathcal{R}T\log (x_{i}P/P^{\circ })$. Reactions can change $N$, so fractions alone do not set the extensive Gibbs free energy. Its composition Hessian is $\mathcal{R}T({\delta }_{ij}/N_i-1/N)$; it is positive semidefinite by the weighted Cauchy-Schwarz inequality.

Repeated Cauchy-Schwarz inequality gives

The Hellinger affinity of dominated probability measures is the integral of the geometric mean of their probability density functions. It lies in $[0,1]$ by Cauchy-Schwarz inequality. The unnormalized squared Hellinger distance is $2-2$ times this affinity; this identity makes the relation with relative entropy particularly simple.

If $f$ is a measurable function into a separable Banach space and has finite second moments under probability measures $\mu ,{\mu }^{\prime }$, then their Bochner integral expected values satisfyHere the Hellinger distance includes the factor $1/2$ in its square. For a proof, use $\nu =(\mu +{\mu }^{\prime })/2$, $p=d\mu /d\nu$, and $q=d{\mu }^{\prime }/d\nu$. Apply the norm bound for a Bochner integral and the Cauchy-Schwarz inequality to $f(p-q)=f(\sqrt{p}-\sqrt{q})(\sqrt{p}+\sqrt{q})$, then use $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$. This explains why stability of a Bayesian posterior in Hellinger distance controls posterior expected values when the relevant second moments remain bounded.

For $(Tg)(v)=\int k(v,w)g(w)dw$, the Cauchy-Schwarz inequality gives $\Vert Tg{\Vert }_2\le \Vert k{\Vert }_{L^2(v,w)}\Vert g{\Vert }_2$. The kernel norm is the Hilbert-Schmidt norm, which bounds the operator norm. Uniform kernel bounds apply also to spatially parametrized operators after integrating over position.

For finite positive measures $\mu ,\nu$, put $\rho =\mu +\nu$. The linear functional $g\mapsto \int gd\nu$ is bounded on real $L^2(\rho )$ by the Cauchy-Schwarz inequality. The Riesz representation theorem gives a measurable $h$ with $\nu (A)={\int }_{A}hd\rho$. Testing indicators of its negative and greater-than-one level sets shows $0\le h\le 1$ almost everywhere. Subtraction gives $\mu (A)={\int }_A(1-h)d\rho$. This constructs the densities without assuming the Radon-Nikodym theorem as a prior result.

For $f\in L^2(0,1)$ and $0<\alpha <1/2$, the integral $g(x)={\int }_0^{1}f(y)|x-y{|}^{-\alpha }dy$ is well defined for every real $x$. Scaling the squared kernel difference over the real line and using the Cauchy-Schwarz inequality give $|g(x+h)-g(x)|\le C_{\alpha }\Vert f{\Vert }_2|h{|}^{1/2-\alpha }$.

If $n$ points and $m$ curves have the property that each pair of distinct points lies on at most $\lambda$ curves, then double counting gives ${\sum }_{\gamma }{k}_{\gamma }(k_{\gamma }-1)\le \lambda n(n-1)$, where $k_{\gamma }=|P\cap \gamma |$. The Cauchy-Schwarz inequality implies $I^2\le mI+\lambda mn(n-1)$ and hence $I\le m+\sqrt{\lambda mn(n-1)}$. The statement needs only this combinatorial multiplicity condition, not algebraicity.

An inner-product space is a vector space equipped with an inner product. The inner product induces the norm $\Vert x\Vert =\sqrt{⟨x,x⟩}$ and hence a metric.

On an oriented even-dimensional Riemannian manifold of strictly positive sectional curvature, a nonconstant closed geodesic has a length-decreasing smooth variation. Its parallel transport fixes the tangent and acts on the odd-dimensional normal space by a special orthogonal group element. An odd-dimensional special orthogonal transformation has a fixed vector, so there is a nonzero periodic parallel normal field $V$. The second variation of geodesic energy is then strictly negative. The Cauchy-Schwarz inequality converts lower energy into strictly lower length because the original geodesic has constant speed. For an embedded geodesic the small variation is a smooth isotopy; otherwise it is a deformation through immersed loops.

Suppose $k_{\tau }$ is a positive-semidefinite kernel for every $\tau$ and its diagonal entries are integrable. Positivity of each $2\times 2$ kernel matrix givesThe Cauchy-Schwarz inequality makes every entry integrable, and integrating each nonnegative finite quadratic form shows that $\int k_{\tau }d\tau$ is again a positive-semidefinite kernel.

For an $L^2$ kernel and an independent sample from a density $f$, the centred kernel density estimator hasIndependence removes cross terms and Tonelli theorem integrates the pointwise variance. Young's convolution inequality ensures $K_h\ast f\in L^2$ even without $f\in L^2$. The Cauchy-Schwarz inequality yields the expected-norm bound $\mathbb{E}\Vert {\hat{f}}_{n,h}-K_h\ast f{\Vert }_2\le \Vert K{\Vert }_2/\sqrt{nh}$.

On $I=(-a,a)$, a first-order Sobolev space function has a continuous representative and obeysBound the average by the Cauchy-Schwarz inequality and average the identity $u(x)-u(y)={\int }_y^x{u}^{\prime }$. This also gives $[u{]}_{C^{0,1/2}}\le \Vert u^{\prime }{\Vert }_2$. A unit constant with the standard Sobolev norm cannot be asserted on arbitrary interval lengths: $u=1$ already disproves it when $2a<1$.

For continuous local martingales, the total-variation process of their quadratic covariation satisfiesIt is the Cauchy-Schwarz inequality for the matrix-valued measure formed by their quadratic variations and quadratic covariation.

For an interval set avoiding one residue at each sieving prime, the Ramanujan sum at the forbidden Chinese remainder theorem residue gives a linear combination of Fourier samples equal to $\mu (d)|S|$. Cauchy-Schwarz inequality gives sample energy at least $|S{|}^2/\phi (d)$ for each squarefree modulus. Summing these energies and applying the analytic large sieve inequality gives the bound. If primes dividing $q$ are excluded, restrict the denominator to $(d,q)=1$; it is uniformly at least a constant times $(\phi (q)/q)\log (2Q)$.

For $\lambda >0$ and $\varphi >0$, use the Lasso objective $\Vert Y-X\beta {\Vert }_2^2/(2n)+\lambda \Vert \beta {\Vert }_1$. Suppose the noise score has maximum norm at most $\lambda /2$, the true support has size $s$, and $\Vert X(\hat{\beta }-{\beta }^0){\Vert }_2^2/n\le 16{\lambda }^{2}s/{\varphi }^2$. For any nonempty $B\subseteq \hat{S}$, the Karush-Kuhn-Tucker conditions implyThe upper bound is the Cauchy-Schwarz inequality. Hence $|B|\le 64{\kappa }_{|B|}^{2}s/{\varphi }^2$. A finite first index $m_{\ast }$ violating this inequality must exceed $|\hat{S}|$. Its minimality also gives $|\hat{S}|\le 64{\kappa }_{m_{\ast }}^{2}s/{\varphi }^2$. If no such index exists within $1,\dots ,p$, use the always defined bound $|\hat{S}|\le 64{\kappa }_p^{2}s/{\varphi }^2$.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

Let $K(\Delta )={\max }_t\#\{r:\Vert {\theta }_r-t\Vert \le \Delta /2\}$. Without a separation assumption,For $0<\Delta \le 1$, each point belongs to at most $K(\Delta )$ integration arcs in the Sobolev–Gallagher inequality. For $\Delta \ge 1$, $K(\Delta )$ is the total number of points, and the Cauchy-Schwarz inequality bound $|S{|}^2\le N\sum |a_n{|}^2$ suffices.

A normalized bipartite pure state of Schmidt rank $r$ has quantum fidelity at most $\sqrt{r/d}$ with a fixed maximally entangled state on $d\otimes d$. Its coefficient matrix has matrix rank $r$ and Hilbert-Schmidt norm one. Its overlap is the trace of that matrix divided by $\sqrt{d}$; trace duality and the Cauchy-Schwarz inequality bound this by $\sqrt{r/d}$. Uniform Schmidt coefficients on $r$ matching basis pairs attain equality.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.