Past exam of the methods of applied functional analysis course of the University of Oxford / 2009 / A / 1

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.