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Transform convention and derivatives. Use the Fourier transform pairFor sufficiently decaying $f,f^{\prime }$ with integrable second derivative, two applications of integration by parts giveThe identity also holds for tempered distributions, which is the setting needed for the Dirac delta function boundary data.

Upper half-plane solution. Take the Fourier transform in $x$ only, writing $\Phi (k,y)=\tilde{\varphi }(k,y)$. Laplace's equation becomes ${\Phi }_{yy}-k^2\Phi =0$. For $k\ne 0$, discard the exponentially growing solution and use $\Phi (k,0)=1$, the Fourier transform of the Dirac delta function. This gives $\Phi (k,y)=e^{-|k|y}$. The value at $k=0$ can be set to its continuous limit one. Pointwise decay of $\varphi$ as $y\to \infty$ does not require this individual transformed value to decay: the total integral in $x$ remains one while the profile spreads.

The inverse Fourier transform now yields, for $y>0$,Evaluate the last integral as the real part of ${\int }_0^{\infty }{e}^{-(y-ix)k}dk=(y-ix{)}^{-1}$:This is the Poisson kernel for the upper half-plane. Its boundary condition is a distributional convergence, rather than an ordinary value at $(0,0)$: for a smooth compactly supported test function $g$,The dominated convergence theorem applies because $g$ is bounded and $(1+t^2{)}^{-1}$ is integrable. Also $0\le \varphi (x,y)\le 1/(\pi y)$, proving the required decay as $y\to \infty$.

Direct verification. Differentiating the rational expression givesTheir sum is zero throughout $y>0$, so the solution is a harmonic function there.

A weak solution is $u\in H_0^1(\Omega )$ such thatLet $M_f=\Vert f{\Vert }_{\infty }$ and define the Nemytskii operator $N(v)=f\circ v$ on $L^2(\Omega )$. It satisfies $\Vert N(v){\Vert }_2\le M_f|\Omega {|}^{1/2}$. It is continuous in $L^2$: from any sequence $v_j\to v$ in $L^2$, every subsequence has a further subsequence converging almost everywhere; continuity and boundedness of $f$, followed by the dominated convergence theorem, give convergence of the corresponding $N(v_j)$ in $L^2$. This subsequence criterion proves continuity of the whole sequence.

For $h\in L^2$, the Lax-Milgram theorem defines the inverse Dirichlet solution map $Sh\in H_0^1$ by $\int \nabla (Sh)\cdot \nabla \phi =\int h\phi$. Testing with $Sh$ and using the Poincaré inequality yieldsThe Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. Therefore $T=S\circ N$ is a nonlinear compact operator and its image lies in the closed $L^2$ ball of radius $R=1+C_P^2{M}_f|\Omega {|}^{1/2}$. This ball is nonempty, closed, bounded and convex. The Schauder fixed-point theorem gives $u=Tu$. Since $Tu\in H_0^1$, its defining identity is exactly the required weak formulation. Thus a weak solution exists. This is the bounded semilinear Dirichlet existence theorem.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

One version of dominated differentiation under the integral sign is the following. Let the parameter range be open. Suppose $f(\cdot ,y)$ is integrable, ${\partial }_{y}f(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable function of $x$ independent of $y$. ThenThis follows by applying the dominated convergence theorem to difference quotients, bounded by that same integrable function through the mean value theorem.

On any compact parameter interval $0<y\le B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin (x^{2}y)|\le B{x}^2$. Its first and second parameter derivatives areTheir absolute values are bounded respectively by $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty )$. Applying the theorem twice proves that $F$ is twice differentiable and givesThis last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt{y}$ yieldsThe expression for $c$ is finite and independent of $y$, which is all that is required.

On any bounded parameter interval $|t|\le T$, the derivative of the integrand has the integrable majorantThe dominated convergence theorem therefore justifies differentiation under the integral sign at every real $t$, givingUse integration by parts in this absolutely convergent integral, with $x{e}^{-x^2}dx=-\frac{1}{2}d(e^{-x^2})$. Both boundary terms vanish, since the sine is zero at $x=0$ and the Gaussian tends to zero at infinity. HenceIt follows that $f^{\prime }=-4t^{3}f$, or equivalently $(e^{t^4}f(t){)}^{\prime }=0$. This identity holds on all of $\mathbb{R}$, including zero. With the supplied Gaussian integral value,The fourth power comes from the printed phase $2x{t}^2$; the parameter has been differentiated with that phase throughout.

The monotone convergence theorem says that if $0\le f_j\uparrow f$ almost everywhere on a measure space, then $\int f_j\uparrow \int f$, allowing infinity. The dominated convergence theorem says that if $g_j\to g$ almost everywhere and $|g_j|\le H$ for an integrable $H$, then $g$ is integrable and $\int g_j\to \int g$.

For $a\ne 0$, the truncated functions $f_a{1}_{[-j,j]}$ increase to $f_a$. Each has the elementary integral $2\arctan (|a|j)/|a|$. Thus monotone convergence givesFor fixed $x$, eventually $|x|\le n\pi /2$ and $n\sin (x/n)\to x$, so $g_n(x)\to 1/(1+x^2)$. Inside the support the supplied inequality yields $n^2{\sin }^2(x/n)\ge 4x^2/{\pi }^2$. Outside the support $g_n=0$. Consequently the global bound isand this majorant is integrable by the already proved formula with $a=2/\pi$. Dominated convergence now gives $\boxed{\int g_n\to \int (1+x^2{)}^{-1}dx=\pi }$. The majorant is independent of $n$, including on the moving support.

Put $G_N={\sum }_{n=1}^N|f_n|$. These are nonnegative measurable functions increasing to $G={\sum }_{n=1}^{\infty }|f_n|$, possibly initially with value infinity. Monotone convergence theorem givesAn infinite value on a set of positive measure would force an infinite integral. Thus $G$ is finite almost everywhere and, after setting it to zero on its null exceptional set if desired, is integrable.

Where $G$ is finite, the real series ${\sum }_n{f}_n$ is absolutely convergent. Define its sum to be $h$ there and zero on the exceptional null set. Measurability follows as an almost-everywhere limit of partial sums, and $|h|\le G$ almost everywhere gives integrability. Every partial sum is dominated by $G$, so dominated convergence theorem yieldsIn fact the convergence is also in the integrable-function norm:

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

We first prove the differentiation theorem. Fix $t\in (0,b)$ and take nonzero $h\to 0$ with $t+h\in (0,b)$. The measurable difference quotientsconverge pointwise to $f_t(x,t)$. The mean value theorem and the assumed derivative bound give $|q_h(x)|\le g(x)$. In particular $f_t(\cdot ,t)$ is measurable as a limit of quotients along a sequence, and its bound by $g$ proves integrability. The dominated convergence theorem, in its parameter-limit form, says that pointwise convergence with a common integrable bound permits convergence of the integrals. Applying it yieldsThe parameter-limit statement follows from the ordinary sequential theorem by applying it to every sequence of nonzero $h$ tending to zero.

For the specific function, $|f(x,t)|\le e^{-x^2}$, which is integrable by the allowed fact. It is differentiable in $t$, withThe last majorant has integral one on $[0,\infty )$, using the substitution $u=x^2$. All three hypotheses therefore hold. The indicator of the parameter domain is one at every point under consideration and does not affect this derivative.

The dominated convergence theorem says that if measurable $f_n\to f$ almost everywhere and $|f_n|\le g$ for an integrable $g$, then $f$ is integrable, $\int |f_n-f|\to 0$, and consequently $\int f_n\to \int f$.

First derive Fatou's lemma from monotone convergence. For nonnegative $u_n$, the measurable functions $v_n={\inf }_{j\ge n}{u}_j$ increase to $\mathrm{liminf}{u}_n$, and $\int v_n\le {\inf }_{j\ge n}\int u_j$. Monotone convergence therefore yields $\int \mathrm{liminf}{u}_n\le \mathrm{liminf}\int u_n$.

Now $f$ is measurable and $|f|\le g$ almost everywhere, so is integrable. Apply Fatou's lemma to $u_n=2g-|f_n-f|\ge 0$, which converges almost everywhere to $2g$:The last integrals are nonnegative, so they tend to zero. Finally $|\int f_n-\int f|\le \int |f_n-f|\to 0$. This argument also works for complex-valued functions.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

The monotone convergence theorem states that for nonnegative measurable functions $f_n\uparrow f$ almost everywhere, $\int f={\lim }_n\int f_n$, allowing the value $+\infty$.

The dominated convergence theorem states that if measurable functions $f_n\to f$ almost everywhere and $|f_n|\le g$ almost everywhere for some integrable $g$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$ as well. Both statements apply on any measure space.

A sufficient differentiation under the integral sign theorem is the following local domination criterion. Suppose $u(t,\cdot )$ is measurable for each $t$, $u(t_0,\cdot )$ is integrable at one point of a neighborhood $J$ of $t_0$, and for almost every $x$ the function $u(\cdot ,x)$ is differentiable on $J$. Assume an integrable function $g_J$ satisfies $|{\partial }_{t}u(t,x)|\le g_J(x)$ for $t\in J$ and almost every $x$. Then $I$ is differentiable on the interior of $J$ andThe mean value theorem bounds the difference quotients by $g_J$, and the dominated convergence theorem passes their pointwise limit through the integral. Applying this criterion locally throughout $(a,b)$ proves differentiability there. One may impose the stronger condition that every $u(t,\cdot )$ is integrable from the outset.

Expand $x^x=e^{x\log x}$. The absolute values of all terms have total integralThe series converges, for instance because $j^{-j}\le 2^{-j}$ for $j\ge 2$. Part (b) applied to the absolute values, followed by subtraction of positive and negative parts, therefore justifies termwise integration. It gives the other Sophomore's dream identity:Equivalently, the partial sums are dominated by $e^{|x\log x|}\le e^{1/e}$, so the dominated convergence theorem gives the same justification.

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give

The optional stopping theorem at a bounded stopping time states that an integrable martingale $Z_n$ satisfies $\mathbb{E}{Z}_T=\mathbb{E}{Z}_0$ whenever $T\le n$ for a fixed finite $n$. Apply this first to $S$ at $T=n\wedge \tau$. The simple symmetric random walk cannot cross an integer endpoint without hitting it, soThe preceding part gives $\tau <\infty$ almost surely, hence $S_{n\wedge \tau }\to S_{\tau }$ almost surely. The constant bound $C=\max \{-a,b\}$ permits the dominated convergence theorem, yielding $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }$ is either $a$ or $b$, solving $ap+b(1-p)=0$ gives its full probability distribution:Now letThis counts the visit at time $0$, and excludes the exit time. Apply the same bounded-time optional stopping theorem to $M_n=|S_n|-L_n$. For each $n$ it givesThe visit counts increase to $L_{\tau }$, so monotone convergence applies to the left. Dominated convergence applies to the right using $|S_{n\wedge \tau }|\le C$. Consequently $L_{\tau }$ is an integrable random variable andThis is the origin case of expected occupation before interval exit for a simple symmetric random walk. The argument obtains integrability of the total occupation count before using any unbounded-time optional stopping for $M$, so it does not presuppose the uniform integrability asked about next. The endpoint probabilities also agree with the standard gambler's ruin calculation.^[ref]

Yes. The preceding calculation established $\mathbb{E}{L}_{\tau }=-2ab/(b-a)<\infty$. For every $n$, the stopped martingale satisfiesHere $H$ is a single nonnegative integrable random variable that bounds the whole family. Thus, for any $K>0$,The last limit follows from the dominated convergence theorem. This is precisely uniform integrability. It also gives $M_{n\wedge \tau }\to M_{\tau }=|S_{\tau }|-L_{\tau }$ with convergence in L1, and therefore $\mathbb{E}{M}_{\tau }=0$, consistently with the preceding answer. A common bound on the individual expected values alone would not have sufficed; the integrable dominating random variable is what supplies control of every tail.