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The weak derivative condition isBoth integrals exist since $u,v$ are locally integrable and the test function has compact support. Derivatives are defined up to almost-everywhere equality. The space $W^{1,p}(\Omega )$ consists of $u\in L^p(\Omega )$ whose first weak derivatives $D_{i}u$ all lie in $L^p(\Omega )$. One standard norm isA sum in the last formula gives an equivalent norm. For restriction to $E$, extend any $\varphi \in C_c^{\infty }(E)$ by zero to $\Omega$. Its support is compactly contained in $E$, so the extension is smooth. The identity over $\Omega$ then reduces exactly to the identity over $E$, proving locality of the weak derivative.

For the jump contribution to a piecewise weak derivative, integration by parts on the two sides of the joining point gives the distributional derivativeThe coefficient is the right trace $1+a$ minus the left trace $2$. If a locally integrable weak derivative existed, locality on each open side would identify it almost everywhere with the displayed regular part. Substituting that into the global test identity would force $(a-1)\varphi (1)=0$ for every test function, hence $a=1$. Equivalently a nonzero Dirac point mass cannot be represented by a locally integrable derivative.

For $a=1$ the point-mass term vanishes, and both the function and its piecewise derivative are bounded on the finite interval. Thus $u\in W^{1,\infty }$ and hence in every listed $W^{1,p}$. Therefore membership holds exactly for $a=1$. Changing a single point value is irrelevant to Sobolev membership; it cannot remove mismatched one-sided traces when $a\ne 1$.

The Lax-Milgram theorem states that a bounded bilinear form $a$ on a real Hilbert space $V$, satisfying $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, has for each $F\in V^{\ast }$ a unique $u\in V$ with $a(u,v)=F(v)$ for all $v$. Moreover $\Vert u{\Vert }_V\le \Vert F\Vert /\alpha$.

For a compact operator $K$ on a Hilbert space, the Fredholm alternative says that $I-\lambda K$ has closed range, finite-dimensional kernel and cokernel of equal dimension, andThus a zero kernel gives unique solvability for every right-hand side; a nonzero kernel gives solvability exactly under the adjoint orthogonality conditions, and then all solutions differ by a kernel vector.

Take $V=H_0^1(\Omega )$. Gårding's inequality supplies constants $a>0$, $b\ge 0$ with $B[v,v]\ge a\Vert v{\Vert }_{H^1}^2-b\Vert v{\Vert }_2^2$. Therefore for $\lambda \le -b$ the shifted form $B_{\lambda }[u,v]=B[u,v]-\lambda (u,v{)}_{L^2}$ is coercive. It is bounded for every fixed $\lambda$, and $v\mapsto (f,v)$ is bounded on $V$. Lax-Milgram gives unique weak solvability for $\lambda \le -\gamma$, with $\gamma =b$. The inequality itself follows, for example, by absorbing the drift cross-term $\Vert b{\Vert }_{\infty }\Vert \nabla v{\Vert }_2\Vert v{\Vert }_2$ using Young's inequality and allowing a lower $L^2$ correction for $c$.

For the compact inverse and adjoint of a coercive elliptic operator, under the additional zero-shift coercivity assumption, define $Gg\in V$ by $B[Gg,v]=(g,v)$ for all $v\in V$, and let $K$ be this solution viewed in $L^2$. The solution operator is linear and bounded into $V$, and the Rellich-Kondrachov compact embedding $V\hookrightarrow L^2$ makes $K$ compact. Rewriting the weak equation as $B[u,v]=(f+\lambda u,v)$ givesConversely an $L^2$ solution of this operator equation lies in $V$, because it equals $K(f+\lambda u)$, and satisfies the weak differential equation.

For constant drift coefficients and zero Dirichlet traces, integration by parts gives $\int b_i{D}_{i}vu=-\int b_{i}v{D}_{i}u$. Symmetry of the principal coefficients therefore makes $B^{\ast }[u,v]=B[v,u]$ the unshifted weak form of the displayed formal adjoint. The shifted adjoint equation requiresThe spectral shift in an adjoint weak formulation is essential: the printed sentence omitting it is false for nonzero $\lambda$. For instance take a smooth ball, $L=-\Delta +1$, any nonzero $u\in C_c^{\infty }(\Omega )$ and $f=-\Delta u+u$. The printed unshifted identity holds, but $(L^{\ast }-\lambda )u=f-\lambda u\ne f$ whenever $\lambda \ne 0$. This counterexample also meets zero-shift coercivity and constant-drift assumptions. The corrected identity above is used for the remainder of the argument.

Let $G_{\ast }g\in V$ solve $B^{\ast }[G_{\ast }g,v]=(g,v)$, which exists by Lax-Milgram because the diagonal coercivity is unchanged. For $p,q\in L^2$,Thus $G_{\ast }$, viewed as an $L^2$ operator, is exactly $K^{\ast }$. Applying this inverse to the corrected shifted adjoint equation givesand the converse follows from its range lying in $V$, just as before.

Finally let the original homogeneous equation have a nonzero solution. Its parameter is nonzero, since the unshifted problem is coercive. Fredholm gives solvability precisely when $(Kf,w)=0$ for every $w\in \ker (I-\lambda K^{\ast })$. Such a $w$ satisfies $K^{\ast }w=w/\lambda$, so $(Kf,w)=(f,w)/\lambda$. Hence the necessary and sufficient condition isWhen this holds, the solutions are an affine translate of the original finite-dimensional homogeneous solution space. The compatibility space is the adjoint kernel; a drift term does not generally permit replacing it by the original kernel.

For the log-cosh obstacle variational inequality, the admissible set is convex: for any $v\in M$, $u+t(v-u)$ remains above the obstacle and in $H_0^1$ for $0\le t\le 1$. Minimality and the right derivative at zero therefore giveThe one-sided sign comes from admissibility, which permits motion toward every $v\in M$ but does not permit an arbitrary negative variation at a contact point.

Now fix $\varphi \in C_c^{\infty }(D)$. Smoothness on $D$ and continuity of the obstacle give a strictly positive minimum of $u-g$ on the compact support of $\varphi$. Hence $u+t\varphi$ is admissible for both signs of sufficiently small $t$. Applying the inequality to both variations yieldsIntegration by parts gives $-\Delta u+\tanh u=0$ distributionally on $D$, and smoothness makes it pointwise. ThusOn the contact set the variational inequality, rather than unrestricted stationarity, remains the governing statement. Convexity of the energy also makes the inequality sufficient for minimization; strict monotonicity implies at most one minimizer, although existence was already stipulated in this part.

A weak solution is a function $u\in H_0^1(\Omega )$ satisfying the weak formulationMultiplication by a test function and integration by parts give this identity from the classical equation. Conversely it is the equation in the sense of distributions, with the zero boundary condition encoded by the zero-boundary Sobolev space. Testing only against $C_0^{\infty }(\Omega )$ is equivalent, by its density in $H_0^1$ and continuity of both sides.

For one coordinate, integration by parts gives ${\int }_{\Omega }({\partial }_{j}u)\phi =-{\int }_{\Omega }u{\partial }_j\phi$: the test function vanishes outside a compact subset of $\Omega$, so there is no boundary contribution. Repeat this operation once for each of the $|\alpha |$ coordinate differentiations in the multi-index $\alpha$. Each operation contributes a factor $-1$, givingAll intermediate functions are continuous on the relevant compact support, since $u\in C^k$ and $|\alpha |\le k$.

An integrable example on the whole real line is the tent functionIt is not differentiable at $-1,0,1$. For a test function $\phi$, split the integral into $(-1,0)$ and $(0,1)$ and use integration by parts. The endpoint terms at $-1,1$ vanish because $u=0$, and those at zero cancel because $u$ is continuous. The result is $\int u{\phi }^{\prime }=-\int (Du)\phi$. Thus the corners obstruct classical differentiability but not this weak derivative.

In dimension one, the Sobolev embedding theorem gives $H^1(\Omega )\hookrightarrow L^{\infty }(\Omega )$. Thereforeand, for fixed $w$,Hence the operator is well-defined and monotone in one dimension.

In every dimension $n\ge 2$ it is not well-defined on all of $H^1\times H^1$. We exhibit $w=u=v\in H^1$ for which the purported pairing $\int w^2|\nabla w{|}^2$ is infinite. Choose a ball compactly contained in $\Omega$, put $r=|x-x_0|$, and multiply each following radial expression by a smooth cutoff function that equals one near $r=0$.

For $n\ge 3$, set $w=r^{-\alpha }$ with $\alpha =(n-2)/3$. The local Sobolev norm is finite becausewhereassince $n-3-4\alpha =(-n-1)/3<-1$.

For $n=2$, take $w=(\log (1/r){)}^{1/3}$ on a sufficiently small ball. With $s=\log (1/r)$, the gradient integral is proportional to ${\int }^{\infty }{s}^{-4/3}ds<\infty$, while the weighted integral is proportional to ${\int }^{\infty }{s}^{-2/3}ds=\infty$. Its $L^2$ integral is finite because the area factor contributes exponential decay in $s$. These singular functions have the indicated weak derivatives; deleting a small central ball and passing to the limit in integration by parts leaves no boundary term.

Thus $w^2\nabla u\cdot \nabla v$ need not be integrable in dimensions at least two. This is the squared Sobolev gradient counterexample. With the additional restriction $w\in L^{\infty }$, the form is well-defined and monotone in any dimension.

Fix $i$, write $\alpha =x_{i-1}$, $\beta =x_i$, and let $E=T-S$ on this interval. The Hermite interpolation conditions give bothTwo applications of integration by parts therefore giveOn this single cubic piece, the same calculation also givesHere $S^{(4)}=0$ inside the interval, and all four endpoint values of $E,E^{\prime }$ are zero. Since $T^{\prime \prime }=E^{\prime \prime }+S^{\prime \prime }$, we obtain the exact identityThis is an integral inequality, with no assertion that the integrand on the left has one sign pointwise. The stronger nodal derivative conditions make the boundary terms vanish on each interval separately; part (i) only needed their cancellation after summing.

The stated Hermite interpolating polynomial exists uniquely in ${\Pi }_{2n+1}$. Indeed, a polynomial in the kernel of a linear map of the value-and-derivative data map has a double root of a polynomial at each of $n+1$ distinct interpolation nodes. A nonzero such polynomial would have polynomial degree at least $2n+2$, which is impossible. The map is therefore an injective function and a linear map between vector spaces of the same finite dimension of a vector space, $2n+2$, and hence is a linear isomorphism. No higher smoothness of the original function $f$ is required to construct it.

An explicit illustration uses interpolation nodes $0,1,2$, values $0,1,0$, and zero derivatives at every interpolation node. The clamped cubic spline is $S(x)=3x^2-2x^3$ on $[0,1]$ and its reflection on $[1,2]$. Its Hermite interpolating polynomial is the quartic polynomial $T(x)=x^2(2-x{)}^2$. On $[0,1]$, $T-S=x^2(1-x{)}^2$, and the corresponding reflected formula holds on $[1,2]$. Direct integration gives $J(S)=24$, $J(T)=128/5$, and $J(T-S)=8/5$, exactly as the energy identity predicts.

Put $E=T-S$. Matching interpolation data and endpoint derivatives givesThe endpoint slope constraints make $S$ a clamped cubic spline. The decisive step in its minimum roughness property of a clamped cubic spline is the orthogonality identityProve it separately on each interval. Two applications of integration by parts, using the vanishing fourth derivative of the cubic piece, giveThe second boundary term is zero because $E$ vanishes at both endpoints of this interval. Sum over $i$. At each interior spline knot, the two values of $S^{\prime \prime }{E}^{\prime }$ agree because both factors are continuous, so the contributions cancel. The remaining outer boundary terms vanish because $E^{\prime }$ is zero at both outer endpoints. This proves the identity.

Expanding the square now gives the stronger exact equalityThe last integral is nonnegative, proving the required inequality. It measures the excess second derivative roughness penalty of $T$ over $S$.

The argument applies directly to twice continuously differentiable competitors. More generally it holds when $T^{\prime }$ is an absolutely continuous function and $T^{\prime \prime }$ is square-integrable, the usual finite-energy class for this second derivative roughness penalty. Mere existence of pointwise second derivatives does not by itself guarantee that all displayed integrals exist; the variational statement is understood for admissible competitors with this regularity.

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

Multiply the equation for $y_i$ by $y_j$ and the equation for $y_j$ by $y_i$, then subtract. The $q$ terms cancel and the derivative terms combine into a Wronskian boundary derivative:Integrating over the interval eliminates the boundary term because both eigenfunctions vanish at both endpoints. Since their eigenvalues differ,This is orthogonality in a weighted inner product in the regular Sturm-Liouville setting; the endpoint values must justify the integration by parts.

The time-independent Schrödinger equation isPut $\alpha =m\omega /\hslash$. For $\psi =e^{-\alpha x^2/2}$, differentiation gives ${\psi }^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\psi$. Substitution cancels the terms in $x^2$ and leaves $E_0={\hslash }^2\alpha /(2m)=\hslash \omega /2$.

The Gaussian integral gives $\int e^{-\alpha x^2}dx=\sqrt{\pi /\alpha }$. Hence a normalized wave function isMultiplication by any constant of modulus one gives the same quantum state. The position probability density is $|{\psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$. Its first moment vanishes by oddness. Integration by parts gives $\int x^2{e}^{-\alpha x^2}dx=(2\alpha {)}^{-1}\int e^{-\alpha x^2}dx$, with the boundary term zero. Therefore

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ where finite. For independent $X,Y$, independence factors the expectation, giving $M_{X+Y}(t)=M_X(t)M_Y(t)$ wherever both factors are finite.

For the given probability density function and $t<1$, put $s=(1-t)x$ to obtainHere $\Gamma (n)=(n-1)!$, obtained by repeated integration by parts. For $t\ge 1$ the defining integral diverges.

Each $Y_i$ has moment-generating function $(1-t{)}^{-1}$, so independence givesThe uniqueness theorem for moment-generating functions applies when the functions are finite on a neighborhood of zero. Hence the sum has the same distribution as $X_n$, a Gamma distribution of shape $n$ and rate one, with the displayed probability density function in the source.

Let $\eta$ be a smooth variation with $\eta (a)=\eta (b)=0$. Assuming the usual continuous second derivatives of $F$, stationarity of the functional under $y+t\eta$ gives its first variationIntegration by parts removes the derivative of $\eta$; its boundary term vanishes because the endpoint values are fixed. Thusfor every such variation. The fundamental lemma of the calculus of variations implies the Euler-Lagrange equation $F_y=(d/dx)F_{y^{\prime }}$.

For arc length, $F=\sqrt{1+(y^{\prime }{)}^2}$, soThe function $s\mapsto s/\sqrt{1+s^2}$ is strictly increasing; hence $y^{\prime }$ is constant. With endpoints $(a,A)$ and $(b,B)$, $a<b$, the extremal isIt is also the global shortest curve: integrating the unit displacement vector dotted with the curve velocity gives the endpoint distance, which is at most the integrated speed by Cauchy-Schwarz inequality; the straight segment attains equality. A vertical segment is described parametrically or after rotating axes, since it cannot be represented by $y(x)$ on a nontrivial horizontal interval.

With fixed endpoint values, a smooth extremal satisfies the Euler-Lagrange equation $F_y-(d/dx)F_{y^{\prime }}=0$. If $F$ has no explicit $x$ dependence, differentiation givesThus $F-y^{\prime }{F}_{y^{\prime }}$ is constant, the Beltrami identity.

For this integrand, $F_y=2y$ and $F_{y^{\prime }}=1+2y^{\prime }$, so the equation is $y^{\prime \prime }=y$. ConsequentlyThe two constants are fixed uniquely by the prescribed values at distinct endpoints. Equivalently, writing these values as $y_1,y_2$,Its first integral is $y^2-y^{\prime 2}$, which is constant. Using only that first integral without the Euler-Lagrange equation could admit spurious constant functions, so the differential equation was used directly.

For any nonzero endpoint-vanishing perturbation $\eta$, the linear terms in $J[y+\eta ]-J[y]$ vanish by the Euler-Lagrange equation and integration by parts; also $\int {\eta }^{\prime }=0$. The remainder isHence the extremal is the unique strict global minimum, and cannot be a maximum.

Take a smooth variation $u+\epsilon \eta$ with $\eta$ vanishing on the boundary; compactly supported test functions suffice for the interior equation. At an extremum, the first variation is zero:Integration by parts removes the derivatives from $\eta$. The boundary terms vanish, givingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationThis is the classical equation for sufficiently smooth admissible extrema; for weaker regularity the first-variation identity supplies its weak form.

For the squared-gradient integrand, the equation is $2(u_{xx}+u_{yy}+u_{zz})=0$, namely the Laplace equation. Put $u_0=ax+by+cz$ and write any admissible function as $u=u_0+v$, with $v=0$ on the boundary. The energy identity isBy the divergence theorem, the middle integral is ${\int }_{\partial V}v(a,b,c)\cdot ndS=0$; equivalently it vanishes by weak integration by parts for zero-trace functions. The final term is nonnegative and vanishes at $v=0$. Thus this is a global minimum, not merely a stationary value:

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.

Take independent smooth variations $y+\epsilon \eta$ and $z+\epsilon \zeta$. Differentiating the functional and using integration by parts gives its first variationFirst choose variations vanishing at the endpoints. To use the stated form of the fundamental lemma of the calculus of variations, take $\eta =(x-a)(b-x)\chi$ with arbitrary smooth $\chi$, and similarly for $\zeta$. The lemma makes $(x-a)(b-x)(F_y-d{F}_{y^{\prime }}/dx)$ vanish, and hence makes the unweighted coefficient vanish in the interior; continuity extends this to the endpoints. Applying the same argument to $z$ gives the Euler-Lagrange equationsIf either endpoint value of $y$ or $z$ is free, the corresponding variation there is arbitrary. The remaining boundary term gives the natural boundary conditions for a free endpoint: $F_{y^{\prime }}=0$ for a free $y$ value and $F_{z^{\prime }}=0$ for a free $z$ value at that endpoint. With both coordinates free at both endpoints, all four conditions apply.

For the specified integrand these equations reduce toand the natural conditions are $y^{\prime }=z^{\prime }=0$ at each free endpoint. Eliminating $z$ givesIts four independent modes give the full general solutionAt zero, $y^{\prime }(0)=C+2D$ and $z^{\prime }(0)=2C+D$. Vanishing forces $C=D=0$, givingAt $x=\pi /6$ the two derivative conditions become $A/2+\sqrt{3}B=0$ and $A+\sqrt{3}B/2=0$. Their coefficient determinant is nonzero, so $A=B=0$. Thus the unique stationary extremal with all four natural boundary conditions is $y=z=0$. The word extremal here means a stationary point of the functional; the indefinite integrand does not require this point to be a minimum.

Use $|\sin (\alpha x)|\le |\alpha |x$ for $x\ge 0$. For the source's functions,The last integral follows by the substitution $u=nx$ and integration by parts. Since ${\sum }_n{n}^{-2}$ converges, (ii) permits summing the integrals and identifies an integrable almost-everywhere sum.

For every $x>0$, the geometric series givesThe functions vanish on $x\le 0$, and the single endpoint does not change the integral. Use the supplied integral of $f_n$ to obtainThe estimates include negative $\alpha$ through its absolute value, and the case $\alpha =0$ gives zero on both sides.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise almost everywhere and the functions are measurable, then $\int h_n\uparrow \int h$, allowing the value infinity.

Interpret the displayed densities as zero outside $(0,1)$, so no undefined logarithm is evaluated there. Each $f_k$ is nonnegative and measurable. With $x=e^{-t}$,where the last equality is integration by parts. This is finite, proving Lebesgue integrability.

The nonnegative partial sums satisfy, for $0<x<1$,They vanish outside the interval as well. Monotone convergence therefore givesThe series is finite by comparison with the convergent series ${\sum }_{j\ge 1}{j}^{-2}$, whose convergence follows from the integral test. Since $g\ge 0$, this finite integral proves absolute Lebesgue integrability as required.