Methods of applied functional analysis course of the University of Oxford (C5.1)

Original solutions to both sections, all six questions and all twenty-five terminal parts of the Methods of Applied Functional Analysis paper sat on 8 June 2010. Source: released Mathematics papers, Part C 2007 11.pdf.

Here uniform ellipticity means that there is a constant $\lambda >0$, independent of $x$, such thatThe assumption that the entries are essentially bounded functions also gives an upper bound $|a(x)\xi |\le \Lambda |\xi |$ with a finite constant $\Lambda$. For this symmetric divergence-form operator, the lower bound makes its energy control the squared Euclidean norm of the gradient.

The space of test functions is $\boxed{C_0^{\infty }(\Omega )=\{\phi \in C^{\infty }(\Omega ):\mathrm{supp}\phi \text{ is a compact subset of }\Omega \}}$. The Sobolev space iswhere $D_{j}u$ denotes a weak derivative, and its Sobolev norm is $\Vert u{\Vert }_{H^1}^2=\Vert u{\Vert }_2^2+{\sum }_j\Vert D_{j}u{\Vert }_2^2$. The zero-boundary Sobolev space is $\boxed{H_0^1(\Omega )={\overline{C_0^{\infty }(\Omega )}}^{H^1}}$. This closure defines the boundary condition even when the boundary has insufficient regularity for a classical boundary value.

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.

Write $a(v,w)={\int }_{\Omega }a\nabla v\cdot \nabla w$ and $A[v]=a(v,v)$. The admissible set is the closed affine subspace $H_g^1=g+H_0^1$. Symmetry of the coefficient matrix givesIf $u$ minimizes the energy, its first variation vanishes: differentiating at $t=0$ gives $a(u,\phi )=0$. Hence $u$ is a weak solution.

Conversely, if $u$ is a weak solution, every admissible $v$ has $v-u\in H_0^1$ and thereforeby uniform ellipticity. This is the Dirichlet principle for the stated symmetric energy.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.

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 $|\alpha |\le k$.

The function $v$ is the weak derivative $D^{\alpha }u$ ifThe integrals exist by local integrability and the compact support of the test function. This is the definition of a distributional derivative represented by a locally integrable function.

The fundamental lemma of the calculus of variations says that if $h\in L_{\mathrm{loc}}^1(\Omega )$ and $\int h\phi =0$ for every $\phi \in C_0^{\infty }(\Omega )$, then $h=0$ almost everywhere.

If $v_1,v_2$ both represent $D^{\alpha }u$, subtracting their defining identities gives $\int (v_1-v_2)\phi =0$ for every test function. The lemma gives $v_1=v_2$ almost everywhere, proving uniqueness. If $u\in C^1$, part (i) gives exactly the defining identity with $v={\partial }_{j}u$. Since this classical derivative is locally integrable, it is the unique weak derivative $D_{j}u$, up to equality almost everywhere.

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.

Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

Write $g=(g_1,\dots ,g_n)$ for the continuous representatives of the weak derivatives. Integrating the preceding identity over $B=B(0,1)$ givesThe first term converges to ${\int }_{B}u$ because $u_{\epsilon }\to u$ in $L^1$. For fixed $x$, all segments in the second term lie in a compact set. Continuity of $g$ implies ${\rho }_{\epsilon }\ast g\to g$ uniformly on that set. Hence every $x$ has the finite limitThe same argument is uniform for $x$ in any fixed compact set: all the segments then lie in a common compact set. Thus $u_{\epsilon }\to U$ locally uniformly, so $U$ is continuous; the local $L^1$ convergence also gives $U=u$ almost everywhere.

Passing to the limit in the segment identity, and taking its endpoints to be $x$ and $x+h$, givesTherefore $U$ is differentiable with continuous gradient $g$: $\boxed{U\in C^1,U=u\text{ almost everywhere},\nabla U=g}$. This is the continuous weak gradient gives a C1 representative criterion.

The real Lax-Milgram theorem states: let $H$ be a Hilbert space and let $a:H\times H\to \mathbb{R}$ be a bounded bilinear form such that $a(v,v)\ge \alpha \Vert v{\Vert }_H^2$ for some $\alpha >0$ and every $v\in H$. For every bounded linear functional $\ell \in H^{\ast }$ there exists a unique $u\in H$ satisfyingSymmetry of $a$ is not required. The complex version replaces bilinearity by sesquilinearity and the lower bound by $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_H^2$.

Introduce coordinates $s=x+y$, $t=x-y$, so $dxdy=\frac{1}{2}dsdt$. For a test function $\phi (x,y)=\Phi (s,t)$,almost everywhere. HenceThe first derivative integrates to zero in $s$, and the second to zero in $t$, using compact support. Thus $\boxed{\mathrm{div}V=0}$ as a weak divergence. Geometrically, the jump across $x+y=0$ is $(1,-1)$, perpendicular to its normal $(1,1)$; there is no distributional flux concentrated on the interface.

A weak solution belongs to the zero-boundary Sobolev space $H_0^1(\Omega )$ and satisfiesThe boundedness of $b$ and the Cauchy-Schwarz inequality make the drift term integrable. This weak formulation moves one derivative from the Laplacian onto the test function, while retaining the first-order drift term.

Give $H_0^1$ the equivalent Hilbert space norm $\Vert u{\Vert }_H=\Vert \nabla u{\Vert }_2$, using the Poincaré inequality. DefineThe Cauchy-Schwarz inequality gives $|a(u,v)|\le (1+C_P\Vert b{\Vert }_{\infty })\Vert u{\Vert }_H\Vert v{\Vert }_H$ and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_H$.

For $\phi \in C_0^{\infty }(\Omega )$, the definition of weak divergence, applied to ${\phi }^2/2$, givesThe left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by test functions and pass to the limit on the left to obtain $\int (b\cdot \nabla u)u\ge 0$. This density argument is essential: local integrability of $\mathrm{div}b$ alone does not justify treating $(\mathrm{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.

Consequently $a(u,u)\ge \Vert u{\Vert }_H^2$, so $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique weak solution with $\boxed{\Vert \nabla u{\Vert }_2\le C_P\Vert f{\Vert }_2}$. The reusable estimate is coercivity from a drift with nonpositive weak divergence.

The Brouwer fixed-point theorem says that every continuous self-map of a closed ball in ${\mathbb{R}}^n$ has a fixed point.

Suppose $g$ has no zero in ${\overline{B}}_R$. Thenis a continuous map from ${\overline{B}}_R$ into its boundary, hence into ${\overline{B}}_R$. The Brouwer fixed-point theorem gives $T(x_0)=x_0$. This forces $|x_0|=R$ andcontradicting the assumed boundary inequality. Therefore $\boxed{g(x_0)=0\text{ for some }|x_0|\le R}$. This is the outward vector field zero theorem.

This is the fixed point theorem for summably Lipschitz iterates. Put $x_n=A^n{x}_0$ and $d=\Vert x_1-x_0\Vert$. For $n\ge 1$,Consequently, for $m>n\ge 1$,Summability makes the tails tend to zero, so $(x_n)$ is a Cauchy sequence. Since $M$ is closed in a Banach space, $x_n\to x\in M$. The $n=1$ hypothesis makes $A$ a Lipschitz map, hence continuous; passing to the limit in $x_{n+1}=A{x}_n$ gives $x=Ax$.

If $y$ is another fixed point, then $\Vert x-y\Vert \le {\alpha }_n\Vert x-y\Vert$ for every $n$. Since ${\alpha }_n\to 0$, eventually ${\alpha }_n<1$ and thus $x=y$. Finally let $m\to \infty$ in the tail estimate:The map itself need not be a contraction mapping; its successive iterates provide the summable error bound.

The space $C([a,b])$ with the supremum norm is a Banach space. Continuity of the kernel makes $K$ a bounded linear operator. Induction on $n$ givesso $A^{n}y-A^{n}z=K^n(y-z)$. The ordered integration region gives the factorial bound for a Volterra iterate:For completeness, if this holds at $n$, then integrating $M{M}^n(s-a{)}^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thusand ${\sum }_{n\ge 1}{\alpha }_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique fixed point. Since $y=Ay$ is precisely the Volterra integral equation, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.

Here $T$ is a nonlinear compact operator: it is continuous and sends bounded subsets of $M$ to relatively compact subsets of $X$. Equivalently, the images of any bounded input sequence have a norm-convergent subsequence.

The Schauder fixed-point theorem states that if $M$ is a nonempty closed bounded convex set in a Banach space and $T:M\to M$ is continuous and compact, then $\boxed{T\text{ has a fixed point in }M}$. Linearity, a contraction estimate, and uniqueness are not assumptions or conclusions of this theorem.

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 displayed nonlinearity is singular at $s=-1$, so it cannot be used directly as the globally bounded function in part (b). On $s\ge 0$, however,Use nonnegative truncation of a semilinear source: define $F(s)=f(\max \{s,0\})$. This is continuous and bounded on all of $\mathbb{R}$, with $F\ge 1/2$. Part (b) gives a weak solution of $-\Delta u=F(u)$, $u\in H_0^1$.

The weak maximum principle in the needed form says that if $u\in H_0^1$ and $\int \nabla u\cdot \nabla \phi \ge 0$ for every nonnegative $\phi \in H_0^1$, then $u\ge 0$ almost everywhere. Here this also follows directly by testing with the negative part $u_{-}=\max \{-u,0\}\in H_0^1$:The Poincaré inequality gives $u_{-}=0$. Hence $u\ge 0$, $F(u)=f(u)$, and the denominator $1+u$ stays at least one almost everywhere. Thus the original equation has a nonnegative weak solution, with bounded right side.

Using the dual pairing between $X^{\ast }$ and $X$, a monotone operator satisfiesA nonlinear coercive operator with respect to $u_0$ satisfiesThe directional derivative is ${\partial }_{v}F(u)={\lim }_{t\to 0}[F(u+tv)-F(u)]/t$, when this finite limit exists through admissible points. At a boundary point of $M$ only an inward direction may be admissible; then one uses the corresponding one-sided limit.

If $A:X\to X^{\ast }$ is monotone and $s<t$, apply its defining inequality at $u+tv$ and $u+sv$. It becomes $(t-s)[\phi (t)-\phi (s)]\ge 0$, so $\phi$ is nondecreasing. Conversely choose $u=y$, $v=x-y$. Nondecrease from zero to one gives $⟨A(x)-A(y),x-y⟩\ge 0$. This proves the line criterion for a monotone operator.

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 $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

The operator is well-defined in every dimension: $|\cos w|\le 1$ givesFor fixed $w$, its monotonicity condition isThus $\cos w\ge 0$ almost everywhere is sufficient. It is also necessary, by the positivity criterion for a scalar gradient form. Indeed, if the bounded coefficient $c=\cos w$ is negative on a set of positive measure, choose a Lebesgue point $x_0$ with $c(x_0)<0$ and a nonconstant compactly supported smooth $\psi$. Set $z_{\epsilon }(x)=\epsilon \psi ((x-x_0)/\epsilon )$, supported inside $\Omega$. Thencontradicting monotonicity. Therefore $\boxed{A(w,\cdot )\text{ is monotone exactly when }\cos w\ge 0\text{ almost everywhere}}$. In particular it is not monotone for every $w$: the admissible constant $w\equiv \pi$ makes it the negative Dirichlet form.

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.

This University of Oxford Part C paper combines Methods of Functional Analysis for PDEs in section A with Fixed point Methods for Nonlinear PDEs in section B. ^[ref] The archive preserves the full editable seven-page paper. Bounded official-source searches found no advertised answers for this exact paper. Question 3 has a missing spectral term in one printed adjoint weak-form assertion; the solution gives a counterexample and the corrected derivation.

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.