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For $t>1$, every forcing point has $\tau <t$. Hence the Green's function givesSince $\int \tau e^{2\tau }d\tau =(\tau /2-1/4)e^{2\tau }$, the remaining integral is $(e^2+1)/4$. ThereforeThis decays at positive infinity.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

Insert the inverse Fourier transforms of $f$ and its complex conjugate, initially under enough decay to justify interchanging integrals:The middle step uses the stated Dirac delta distribution identity, and the last uses its sampling property. This is the Parseval identity for the chosen normalization. Approximation extends the result to square-integrable functions, including the function used below.

The squared norm of the original function isApplying the Parseval identity to part (c), using the evenness of the squared transform, givesThus the last integral equals ${\pi }^2/8$. Now set $u=\pi k/2$. Sincewe obtainThe apparent singularity at $u=\pi /2$ is removable: the continuous value of the integrand there is $1/4$. No principal-value interpretation is needed.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

This is an isoperimetric constraint, imposed on the whole path rather than separately at each time. Introduce a constant Lagrange multiplier $\lambda$ and replace the variational integrand by $\tilde{f}=f+\lambda u$. The modified Euler-Lagrange equations areSolve these together with the fixed endpoint conditions, treating $\lambda$ as an unknown constant, then choose its value so thatThe multiplier is constant because there is one scalar integral constraint. A condition imposed at every time would instead call for a time-dependent multiplier. As usual, this multiplier procedure assumes the constraint is regular, so its first variation does not vanish identically on all admissible variations.

Impose $y=x$ before varying. Writing $\delta ={\delta }_1+{\delta }_2$, the restricted integrand isIts middle term is the total derivative of $\alpha ({\gamma }_1+{\gamma }_2)(x+\beta x^2/2)$, so it can be dropped for fixed endpoints by part (b). The remaining Euler-Lagrange equation is $\delta \ddot{x}=\alpha \beta$. Integrating twice and imposing $x(0)=0$, $x(T)=S$ givesThis is the unique global minimizer among paths on the specified line. To prove it, add any admissible variation $\eta$ with $\eta (0)=\eta (T)=0$. The change in fuel, after dropping the fixed boundary term, isThe first integral vanishes by integration by parts and the equation for $x$. Equality forces $\eta =0$. No monotonicity or restriction to the segment between endpoints was imposed in the question; the minimizing coordinate can initially decrease for sufficiently large $T$.

For $n\ne 0$, split the Green's function integral at $s=x$. Writing $\ell (x)=x^{-n}-2^n$ and $r(x)=x^{-n}-1$ givesFor $n\ne -1$, the two integrals areSubstitution and simplification yieldAt $x=1$ both contributions vanish. At $x=1/2$ the bracket is $-1/2+(2^n-1)/[2(2^n-1)]=0$. The differential operator kills $x^{-n}-1$ and sends $x/(n+1)$ to $1/x$, so this also verifies the inhomogeneous equation.

The exceptional integer is $\boxed{n=-1}$, because then the integrals contain $s^{-1}$. In this caseand the Green's function integral becomesHere $L=d^2/d{x}^2$, and $u^{\prime \prime }=1/x$. Explicitly $u(1)=0$ and $u(1/2)=-(\log 2)/2+(\log 2)/2=0$. Thus both exceptional and nonexceptional formulas satisfy the required boundary conditions.

Assume first that $f$ and $g$ are integrable functions, so that Fubini's theorem justifies changing the order of the two integrals. In the defining Fourier transform of the convolution, put $y=x-x^{\prime }$:This proves the convolution theorem with the stated normalization:

Direct evaluation of the Fourier transform gives, for $k\ne 0$,At $k=0$ the defining integral is the interval length, so $\boxed{\tilde{g}(0)=2}$. This also equals the continuous limit of the displayed formula.

The Translation property of the Fourier transform givesFor the remaining integral, the same Translation property of the Fourier transform identifies $e^{20i{k}^{\prime }}\tilde{g}(k^{\prime })$ as the Fourier transform of $g(x+20)$. By the product form of the convolution theorem, with inverse normalization $1/(2\pi )$,But $g(x+20)$ is supported in $[-21,-19]$, whereas the four translated triangular functions making up $r$ are all supported in $[-8,8]$. Their product is identically zero. Therefore the requested integral is $\boxed{0}$ for every real $k$.

Write the equation as $(L+\beta )y=-\cos x/x$ with the positive Sturm-Liouville operator from part (a). A Sturm-Liouville eigenfunction expansion therefore gives $a_n=-f_n/(4n^2+\beta )$, where weighted orthogonal projection givesHere the product-to-sum formulas reduce the integral to half the sum of the integrals of $\sin ((2n+1)x)$ and $\sin ((2n-1)x)$. ConsequentlyAt $x=0$, use the continuous limits of the terms. The coefficients are $O(n^{-3})$, so the resulting endpoint series converges. Completeness is the ordinary Fourier sine series completeness after $v=xy$. For an independent check, solving $v^{\prime \prime }-\beta v=\cos x$ with zero endpoint values givesIts numerator vanishes at zero, and it has the same orthogonal projection coefficients. Positivity of $L+\beta$ proves uniqueness.

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 $a\ne 1$.

Weak convergence $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^{\ast }$. Weak-star convergence $T_j\stackrel{\ast }{\rightharpoonup }T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The Sequential Banach-Alaoglu theorem for a separable predual says that every norm-bounded sequence in $X^{\ast }$ has a weak-star convergent subsequence. Separability is what turns the bounded weak-star compactness statement into this sequential form.

For the weak-star limit of periodic rescalings, apply it with $X=L^1(\mathbb{R})$, which is separable, and $X^{\ast }=L^{\infty }(\mathbb{R})$. The rescaled functions have uniform essential-supremum bound $\Vert \theta {\Vert }_{\infty }$. Let $q=\theta -\bar{\theta }$ and $F(t)={\int }_0^{t}q(s)ds$. Since $q$ is one-periodic with zero mean, $F$ is a bounded one-periodic absolutely continuous primitive. Thus for every finite interval,Every weak-star subsequential limit $h$ therefore satisfies ${\int }_a^b(h-\bar{\theta })=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and set $v=(h-\bar{\theta })1_{[-R,R]}$. This is in $L^1(\mathbb{R})$, and its integral over any interval is zero, since intersecting with $[-R,R]$ again gives an interval. Hence $v=0$ almost everywhere. Taking integer $R\to \infty$ yields $h=\bar{\theta }$ almost everywhere on the whole line.

Finally, if the full sequence did not converge weak-star to this constant, some $g\in L^1$ and some subsequence would have its pairing with $\theta (jx)-\bar{\theta }$ bounded away from zero. Sequential compactness gives a further weak-star convergent subsequence, whose limit must be the constant just identified, a contradiction. ThereforeThis is averaging against integrable test functions, not a claim of pointwise or strong norm convergence.

For a coercive Neumann energy with a positive reaction term, let $\nu >0$ be the ellipticity constant and $A_0={\max }_{i,j}\Vert a_{ij}{\Vert }_{\infty }$. Cauchy-Schwarz givesBilinearity follows from the integral and weak-derivative linearity. Ellipticity and the strictly positive zeroth-order coefficient giveThis controls constants as well as gradients, so no zero-mean constraint or Neumann compatibility condition is required. Symmetry of the coefficients makes $B$ symmetric.

Write $F(v)={\int }_{\Omega }fv$. This is a bounded linear functional on $H^1$, with $|F(v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Consequentlyas the norm tends to infinity. The infimum is finite, and every minimizing sequence is bounded. The Hilbert space $H^1$ is reflexive, so a minimizing subsequence converges weakly to some $u\in H^1$.

To verify lower semicontinuity directly, put $w_j=v_j-u$. Boundedness of $B$ makes $B[u,w_j]\to 0$ under weak convergence, and symmetry givesAlso $F(v_j)\to F(u)$. Therefore $I(u)\le \mathrm{liminf}I(v_j)$, so the absolute minimum is attained.

For every $h\in H^1$ the real function $t\mapsto I(u+th)$ has a minimum at zero. Its derivative there is $B[u,h]-F(h)$, yielding the required weak equation for every $h$. If two weak solutions exist, their difference $z$ satisfies $B[z,h]=0$ for all $h$; choose $h=z$ and use coercivity to obtain $z=0$. In fact the weak equation also gives the exact energy identityThus the minimizer and weak solution are both unique. The natural boundary condition is encoded by testing in the whole $H^1$ space, without needing pointwise conormal derivatives for rough coefficients.

One form of Schauder's theorem says that if $M$ is a nonempty closed bounded convex subset of a Banach space and $T:M\to M$ is continuous with relatively compact image, then $T$ has a fixed point.

For a Schauder solution of a small Hammerstein integral equation, take $X=C([0,1])$ with the supremum norm and $M=\{u:\Vert u{\Vert }_{\infty }\le R\}$. This set is nonempty, closed, bounded and convex. PutBoth maxima are finite by continuity on compact sets. Define $(T_{\mu }u)(x)=\mu {\int }_0^{1}G(x,y)f(y,u(y))dy$. Continuity and uniform continuity of the kernel give $T_{\mu }u\in X$, and for $0<\mu \le {\mu }_0$,so it is a self-map. This choice of ${\mu }_0$ also handles the case when either maximum is zero without division by zero.

If $u_n\to u$ uniformly in $M$, uniform continuity of $f$ on $[0,1]\times [-R,R]$ makes $f(\cdot ,u_n(\cdot ))\to f(\cdot ,u(\cdot ))$ uniformly. The integral estimate then proves continuity of $T_{\mu }$ in the supremum norm. Moreoverwhich tends to zero with $|x-x^{\prime }|$ uniformly in $u\in M$. Thus the image is uniformly bounded and equicontinuous. Arzelà-Ascoli states that such a family on a compact interval has compact closure in the supremum norm. Hence $T_{\mu }(M)$ is relatively compact, and Schauder gives a fixed point for each $\mu$ in the specified interval. Therefore a continuous solution with supremum norm at most $R$ exists. The argument imposes no Lipschitz hypothesis on $f$ and makes no uniqueness assertion.

For the log-cosh energy on a Sobolev space, the elementary inequalities $0\le \log \cosh t\le t^2/2$ follow by integrating $|\tanh t|\le |t|$ from zero. Thusso the energy is well-defined on $H^1$. For a fixed $v$, its proposed derivative is linear in the direction $h$ and bounded, because $|\tanh v|\le |v|$ givesTo justify differentiation, use $(\log \cosh t{)}^{\prime \prime }={\mathrm{sech}}^{2}t\in (0,1]$. Taylor's integral remainder pointwise, combined with the exact quadratic gradient expansion, givesEvery term is integrable by the displayed bounds. The remainder is quadratic in the norm, so the functional is even Fréchet differentiable, and in particular Gâteaux differentiable with the required derivative. This also proves that $A(v)$ belongs to the stated dual space.

For $v,w\in H^1$,Both terms are nonnegative, and the integrand in the second is strictly positive wherever $v\ne w$, since $\tanh$ is strictly increasing. Distinct Sobolev elements differ on a set of positive measure, so the expression is positive. Hence $A$ is strictly monotone.

This gives strict monotonicity without coercivity on H1: it is not coercive on $H^1$. For constant functions $v_t=t$ with $t\to +\infty$, the norm is $t|\Omega {|}^{1/2}$ andwhich is finite. In contrast it is coercive on $H_0^1$: if $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, thenDividing by the norm proves growth to infinity. On this smaller space the operator is understood by restricting its dual action to zero-trace test functions. Removing the constant directions by the boundary condition is what restores coercivity.

Write $b_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The given limiting identity and the Cauchy coefficient formula from part (a) give the boundary contour integralThe contour passes through $z=1$, so its value here is an improper integral using the one-sided branch values. Parametrize its open arc by $z=e^{i\theta }$, $0<\theta <2\pi$. Substitution of the analytic branch of a square root calculated in part (b) givesThe factor $e^{-in\theta }$ comes from cancelling $e^{i\theta }$ in $dz$ against $z^{n+1}$; the extra $e^{-i\theta /4}$ comes from the chosen branch. Keeping both phases prevents a change in the half-integer frequency.

Now put $t=\theta /2$. ThenDividing by $i-1$ givesTaking the real part therefore provesAll these improper integrals converge absolutely: the numerators are bounded, while $(\sin t{)}^{-1/2}$ is of order $t^{-1/2}$ at zero and $(\pi -t{)}^{-1/2}$ at $\pi$. Hence the substitutions and passage to the real part are legitimate. The imaginary part also shows that the corresponding sine integral has the same positive value, which checks the phase calculation.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.