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If ${\sup }_n\mathbb{E}[X_n^2]<\infty$, then $(X_n)$ is uniformly integrable. Indeed, the Cauchy-Schwarz inequality and Markov inequality give

For complex numbers $|u_n|\le 1$, $0\le n<N$, and $1\le H\le N$, put $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. ThenExtend the sequence by zero, count each summand in $H$ consecutive windows, and apply the Cauchy-Schwarz inequality to their sum. Expansion of the squared window sums gives the stated coefficients. Fixing $H$ and making each normalized correlation tend to zero bounds the limiting squared average by $1/H$; then let $H$ tend to infinity. This is the scalar finite form underlying the Van der Corput lemma.

For a bounded sequence $(u_n)$ in a Hilbert space, put $c_h={\mathrm{limsup}}_{N\to \infty }|N^{-1}{\sum }_{n=0}^{N-h-1}⟨u_{n+h},u_n⟩|$. If $H^{-1}{\sum }_{h=1}^H{c}_h\to 0$, then $N^{-1}{\sum }_{n=0}^{N-1}{u}_n\to 0$ in norm. Since the $c_h$ are bounded, it suffices that $(c_h)$ has convergence in density of a sequence to zero. The proof applies the Cauchy-Schwarz inequality to an average of shifted sums.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.

For $\overline{w{a}^{\prime }}=-{\eta }_t{B}_{0}f$, $⟨f⟩=1$ and positive diffusivities, the Zeldovich planar flux balance gives $G=({\eta }_t/\eta )B_0^2[1-({\eta }_t/\eta )\mathrm{Var}(f)]$, where $G=⟨|\nabla a^{\prime }{|}^2⟩$. The Cauchy-Schwarz inequality and a Poincaré inequality $⟨a^{\prime 2}⟩\le d^{2}G/{\pi }^2$ give ${\eta }_t^2\mathrm{Var}(f)\le ⟨w^2⟩d^{2}s(1-s)/{\pi }^2$, with $s=({\eta }_t/\eta )\mathrm{Var}(f)\in [0,1]$. Maximizing $s(1-s)$ gives the displayed bound. Large turbulent-to-microscopic diffusivity therefore requires $f$ to be close to one in mean square, without guaranteeing pointwise uniformity.

Put $Q={\sum }_i{q}_i$, $M={\sum }_i{q}_i{\mu }_i$, and $A={\sum }_i{q}_i\mathbb{E}{Y}_i^2$. A compound Poisson distribution with count mean $Q$ and severity law ${\sum }_i(q_i/Q)F_i$ has variance $V_{\mathrm{P}}=A$. A compound binomial distribution with $n$ trials, probability $Q/n$, and that same severity law has variance $V_{\mathrm{B}}=A-M^2/n$. The individual-policy variance is $V=A-{\sum }_i(q_i{\mu }_i{)}^2$. The Cauchy-Schwarz inequality gives ${\sum }_i(q_i{\mu }_i{)}^2\ge M^2/n$, proving the ordering. Equality of the first two variances requires constant $q_i{\mu }_i$; it need not imply equality of the distributions. If all probabilities and severity laws are identical, the binomial compound approximation is exactly the individual model.

For a unital represented C-star algebra $A\subseteq \mathcal{B}(H)$, the displayed functional is a state. It is linear under the first-argument-linear inner-product convention, has $x_{\eta }(1)=1$, and is positive because $x_{\eta }(a^{\ast }a)=\Vert a\eta {\Vert }^2\ge 0$. The Cauchy-Schwarz inequality gives $|x_{\eta }(a)|\le \Vert a\Vert$, while evaluation at the unit gives equality of its norm with one. Its GNS null left ideal consists exactly of the operators annihilating $\eta$.

By orthogonality of integer Fourier modes, this integral counts pairs of $k$-tuples with the same first $r$ power sums. Diagonal pairs give $J_{k,r}(Z)\gg Z^k$. There are $O_{k,r}(Z^{r(r+1)/2})$ possible moment vectors; the Cauchy-Schwarz inequality gives $J_{k,r}(Z)\gg Z^{2k-r(r+1)/2}$. Upper bounds measure how much arithmetic coincidence remains beyond these necessary contributions, and enter the Vinogradov mean-value method for a bilinear exponential sum.

Expand moments of the inner sum and group equal power-sum differences. Their multiplicities are bounded by $J_{k,r}(Z)$ through the Cauchy-Schwarz inequality. Two applications of the Holder inequality yield a bound for $|U{|}^{4k^2}$ containing $Z^{8k^2-4k}{J}_{k,r}(Z{)}^2$ and a product of short geometric-sum bounds over the moment differences. Thus a sharp mean-value estimate combines with rational approximation or spacing of the coefficients to prove cancellation. When all coefficients are integers, $U=\lfloor Z{\rfloor }^2$, so mean-value estimates alone cannot force cancellation.

If $\square u=F$ with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, then $\Vert \partial u(t){\Vert }_2\le \Vert \partial u(0){\Vert }_2+{\int }_0^t\Vert F(s){\Vert }_{2}ds$, where $\partial u=(u_t,\nabla u)$. Test the wave equation against $u_t$, use integration by parts, and apply the Cauchy-Schwarz inequality. The same energy estimate applies to a finite system of wave equations.

At design sites $i/n$, let ${\tilde{Y}}_n$ be the step function taking response $Y_i$ on $((i-1)/n,i/n]$. Project it onto an interval-adapted wavelet basis through resolution $J$. Each coefficient is ${\hat{c}}_{\lambda }={\sum }_i{Y}_i{\int }_{I_i}{b}_{\lambda }(t)dt$. Its expectation is the corresponding coefficient of the step function of sampled regression means. For a Haar wavelet basis, dyadic sample size and resolution at most the sampling scale, this is exactly $n^{-1}{\sum }_i{Y}_i{b}_{\lambda }(i/n)$ with right-closed pointwise versions. Independent errors of variance ${\sigma }^2$ give coefficient variance at most ${\sigma }^2/n$: Cauchy-Schwarz inequality on each sampling cell yields ${\sum }_i({\int }_{I_i}{b}_{\lambda }{)}^2\le \Vert b_{\lambda }{\Vert }_2^2/n$.

Let $E_q={\max }_{(a,q)=1}|\pi (x;q,a)-\mathrm{Li}(x)/\varphi (q)|$. For $Q\le x^{1-\eta }$, the Brun–Titchmarsh theorem and Cauchy-Schwarz inequality giveWrite each summand as $\sqrt{E_q}{3}^{\omega (q)}\sqrt{E_q}$, then use $E_q{\ll }_{\eta }x/(\varphi (q)\log x)$.

For $M>0$ with integral one, the Cauchy-Schwarz inequality makes the mass functional bounded on the weighted Hilbert space. The vector $M$ has norm one, so the velocity-reset projection is the orthogonal projection onto its span. Consequently $L=\Pi -I$ is bounded and self-adjoint. This Hilbert geometry differs from the norm-one positive reset projection on phase-space $L^1$.

For two samples of sizes $n_1,n_2>1$, write $a=s_1^2/n_1$ and $b=s_2^2/n_2$, with $a+b>0$. The reference statistical degrees of freedom in Welch's t-test satisfyIndeed the Cauchy-Schwarz inequality applied to $(a/\sqrt{n_1-1},b/\sqrt{n_2-1})$ and $(\sqrt{n_1-1},\sqrt{n_2-1})$ gives the inequality after division. Thus an independent two-sample Welch's t-test on $N$ complete observations cannot have reference statistical degrees of freedom exceeding $N-2$. The exact pooled two-sample t statistic also uses $N-2$, under its stronger equal-variance assumptions.

Taking a finite difference lowers the degree of a polynomial phase. For $S(\theta )={\sum }_{x=1}^{n}e(\theta P(x))$, expand $|S{|}^2$ by shifts $h$ and apply the Cauchy-Schwarz inequality to those shift sums. This gives $|S{|}^4\le 2n{\sum }_{h,l}{\sum }_{x\in I_{h,l}}e(\theta {\Delta }_h{\Delta }_{l}P(x))$, where $I_{h,l}$ enforces that $x,x+h,x+l,x+h+l$ lie in $[1,n]$. Although individual terms can be complex, the entire sum is real and nonnegative because it equals a sum of squares. For $P(x)=x^3$, the phase is $3hl(2x+h+l)$.

For a finite positive Borel measure on the circle, averaged squared Fourier coefficients tend to the sum of squares of its atomic masses. Express each squared coefficient as a double integral and average the circle character. The averaged character is bounded by one, tends to one on the diagonal and to zero elsewhere, so the dominated convergence theorem gives the diagonal mass. In particular an atomless measure has vanishing averaged absolute coefficients by the Cauchy-Schwarz inequality.