Past exam of the methods of applied functional analysis course of the University of Oxford / 2009 / A

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.

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.