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Asymptotic regularization stops the gradient flow $x^{\prime }=-A^{\ast }(Ax-f)$ at $t=1/\alpha$, starting from zero. With the singular system of a compact operator convention $A{v}_j={\sigma }_j{u}_j$, its spectral filter givesThe scalar coefficient solves $c_j^{\prime }=-{\sigma }_j^2{c}_j+{\sigma }_j⟨f,u_j⟩$ with zero initial value. Since $1-e^{-s}\le \min (s,1)$ for $s\ge 0$, the operator norm is at most ${\alpha }^{-1/2}$. On the domain of the Moore–Penrose inverse of an operator, dominated convergence theorem of the squared spectral coefficients proves $R_{\alpha }f\to A^{†}f$. The noise-bias decomposition for linear regularization then gives noisy-data convergence when $\delta /\sqrt{\alpha }\to 0$.

On a finite measure space, if measurable functions are uniformly bounded and converge pointwise, then their integrals converge. This is the dominated convergence theorem with a constant dominating function.

For standard Brownian motion started at zero, $H_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely for $a\ge 0$ andThe Brownian reflection principle proves finiteness. For $u\ge 0$, the Exponential martingale for Brownian motion stopped at $H_a\wedge t$ is bounded by $e^{ua}$. The dominated convergence theorem gives $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$, and taking $u=\sqrt{2\lambda }$ gives the transform.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

The integral $I(x)={\int }_0^{\infty }{e}^{xt-t^3/3}dt$ is a rescaled Scorer Hi function. It has the three useful limitsFor the negative-argument limit, scale $t=s/M$ and use the dominated convergence theorem. At zero, substitute $s=t^3/3$ and use the Gamma integral. For the positive-argument limit, the exponent has its maximum at $t=\sqrt{M}$, with second derivative $-2\sqrt{M}$; Laplace's method gives the displayed factor. These estimates describe a cubic endpoint-to-saddle transition.

For a random walk with independent and identically distributed random variables as increments and finite moment-generating function at $\theta$, put $\kappa (\theta )=\log \mathbb{E}{e}^{\theta X_1}$. Then $e^{\theta S_n-n\kappa (\theta )}$ is a martingale by conditioning on the next independent increment. In particular, if $r>0$ and $\mathbb{E}{r}^{X_1}=1$, then $r^{S_n}$ is a martingale. Stopping at a finite interval exit can make this process bounded, allowing the dominated convergence theorem to justify passage from bounded stopping times to the exit time.

Let $f$ be compactly supported and continuous, let $g,\hat{g}\in L^1(\mathbb{R})$, and use $\hat{g}(t)=\int g(x)e^{itx}dx$. If $g(0)\ne 0$, thenThis follows by Fubini's theorem, a change of variables, Fourier inversion theorem, and dominated convergence theorem. It gives a direct proof of the uniqueness theorem for characteristic functions because the right side depends only on ${\phi }_Z$.

If $u\in W_{\mathrm{loc}}^{1,2}$, then $Du=0$ almost everywhere on every level set $\{u=t\}$. Choose smooth truncations ${\Phi }_{\epsilon }(s)$ with $|{\Phi }_{\epsilon }|\le 2\epsilon$, derivative one near zero, bounded derivative, and derivative zero outside $(-2\epsilon ,2\epsilon )$. The Sobolev chain rule gives $D{\Phi }_{\epsilon }(u-t)\to 1_{\{u=t\}}Du$ in local $L^2$ by dominated convergence theorem, while ${\Phi }_{\epsilon }(u-t)\to 0$. The limiting weak derivative is therefore zero.

An integrable discrete-time local martingale is a true martingale. Its stopped increments have the form $1_{\{\tau \ge t\}}(X_t-X_{t-1})$, and the dominated convergence theorem removes a localizing sequence from the conditional increment identity.

A measurable function $M$ is an integrable envelope of a function class for a class $\mathcal{H}$ under a probability law $P$ if $|h(x)|\le M(x)$ for every $h\in \mathcal{H}$ and $PM<\infty$. Such an envelope permits the dominated convergence theorem to control shrinking function brackets.

On a finite measure space, the convergence in the Birkhoff ergodic theorem also holds in $L^1$. Prove it first for bounded truncations by the dominated convergence theorem, then use the $L^1$ contraction $\Vert A_{n}h{\Vert }_1\le \Vert h{\Vert }_1$ and Fatou lemma to remove the truncation.

A martingale satisfying ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$, for $1<p<\infty$, has almost sure convergence and convergence in Lp to a limit $M_{\infty }\in L^p$. Also $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. The Doob Lp maximal inequality makes ${\sup }_n|M_n|$ integrable to the power $p$; the Martingale convergence theorem gives the almost-sure limit, and the dominated convergence theorem gives convergence in the Lp norm. Boundedness in $L^1$ alone does not imply convergence in L1.

If $f\in L^1(\mathbb{R})$ and $x^{k}f\in L^1(\mathbb{R})$ for $0\le k\le r$, then its Fourier transform is $r$ times continuously differentiable and the displayed formula holds. The dominated convergence theorem permits each derivative under the integral. Thus $r+1$ vanishing moments give a zero Taylor polynomial through degree $r$ at the origin. Unweighted $L^1$ membership only guarantees a continuous function as the Fourier transform.

A nonnegative continuous multiplier $k$ defines a contraction semigroup on the space of continuous functions vanishing at infinity and on $L^2(\mu )$ by $P_{t}f=e^{-tk}f$. Strong continuity follows from uniform convergence on compact sets and small tails in $C_0$, or the dominated convergence theorem in $L^2$. The multiplier itself may be unbounded.