Past exam of the distribution theory course of the University of Oxford / 2023 / 2

Convergence ${\varphi }_j\to \varphi$ in the space of test functions means that all supports lie in one compact set $K\subset \mathbb{R}$ and $\Vert {\varphi }_j^{(r)}-{\varphi }^{(r)}{\Vert }_{\infty }\to 0$ for every integer $r\ge 0$. An eventual common compact support is equivalent, since finitely many initial supports can be added to $K$.

Write the standard mollifier as ${\rho }_{\epsilon }(x)={\epsilon }^{-1}\rho (x/\epsilon )$, where $\rho \in \mathcal{D}(\mathbb{R})$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\epsilon \le 1$, the convolution ${\rho }_{\epsilon }\ast \varphi$ has support in the fixed compact set $\mathrm{supp}\varphi +[-1,1]$. The given convolution derivative identity yieldsEach ${\varphi }^{(r)}$ is uniformly continuous; more explicitly, the mean value theorem bounds the absolute value by $\epsilon \Vert {\varphi }^{(r+1)}{\Vert }_{\infty }\int |y|\rho (y)dy$, uniformly in $x$. Thus

If ${\varphi }_j\to 0$ in the space of test functions and $K$ is a common compact support, then $\Vert {\varphi }_j{\Vert }_{L^1}\le |K|\Vert {\varphi }_j{\Vert }_{\infty }\to 0$, where $|K|$ is its Lebesgue measure.

The mollifiers themselves have no limit in the L1 space. If ${\rho }_{\epsilon }\to g$ in that norm, then for each $\delta >0$ the integral of $|g|$ on $\{|x|>\delta \}$ is zero, because ${\rho }_{\epsilon }$ vanishes there for $\epsilon <\delta$. Hence $g=0$ almost everywhere. This contradicts $\Vert {\rho }_{\epsilon }{\Vert }_{L^1}=1$. Therefore $\boxed{{\rho }_{\epsilon }\text{ does not converge in }{L}^1(\mathbb{R})}$.

The weak convergence of distributions $u_j\to u$ meansFor $\theta \in \mathcal{D}(\mathbb{R})$, its convolution with a distribution is the smooth functionThe subscript records the variable on which the distribution acts. With $\check{\theta }(x)=\theta (-x)$, its regular distribution satisfies the useful pairing identityIndeed integrating $\theta (x-y)\varphi (x)$ in $x$ gives $(\check{\theta }\ast \varphi )(y)$; continuity of the distribution justifies moving the pairing through that integral, by approximation with sums of test functions in one common compact support.

Part (i), applied to the reflected mollifier, gives ${\check{\rho }}_{\epsilon }\ast \varphi \to \varphi$ in the space of test functions. Continuity of $u$ therefore provesReflection is harmless even if the chosen standard mollifier is not even.

For the mollifiers themselves, a change of variable givesThus $\boxed{{\rho }_{\epsilon }\to {\delta }_0\text{ in }{\mathcal{D}}^{\prime }(\mathbb{R})}$, with limit the Dirac delta distribution, despite the failure of convergence in the L1 space.

The L1-normalized dilation of a distribution isFor a regular distribution this is the function $u_{\epsilon }(x)={\epsilon }^{-1}u(x/\epsilon )$. In terms of the wiki's dilation of a distribution convention, $u_{\epsilon }={\epsilon }^{-1}{\delta }_{1/\epsilon }u$; the normalization preserves the L1 space norm.

If $u\in L^1(\mathbb{R})$, the dominated convergence theorem, with dominating integrable function $|u(y)|\Vert \varphi {\Vert }_{\infty }$, givesConsequentlyAn arbitrary distribution need not have this behavior. Take the constant regular distribution $u=1$. Then $u_{\epsilon }={\epsilon }^{-1}$, so for a test function with integral one,Hence $\boxed{u=1\text{ gives no distributional limit}}$.

Put $A={\int }_0^{1}f$. ThenThe theorem on differentiation of an indefinite Lebesgue integral makes this an absolutely continuous function on $[0,1]$, withBoth $A-f$ and $Rf$ belong to the L1 space: for example, $|Rf(x)|\le |A|+\Vert f{\Vert }_{L^1}\le 2\Vert f{\Vert }_{L^1}$. The ordinary derivative of an absolutely continuous function is its weak derivative, as integration by parts against a test function shows. Therefore $\boxed{Rf\in W^{1,1}(0,1)}$, the indicated Sobolev space.

Using $(Rf{)}^{\prime }={\int }_0^{1}f-f$ and integration by parts, with $\varphi$ a test function, givesFor each $k\ge 1$, apply the same identity to $R^{k}f$ and the test function ${\varphi }^{(k)}$. Since ${\int }_0^1{\varphi }^{(k)}=0$, it givesStarting from $k=1$ and iterating provesThe intermediate means disappear because derivatives of a test function have zero integral; no assumption that $R^{k}f$ has zero mean is needed.

For the bound, writeThe triangle inequality and Fubini's theorem implyThus $\boxed{c=\frac{1}{2}\text{ is valid}}$, and every iterate $R^{k}f$ is in the L1 space.

To repeat integration by parts across whole periods, choose periodic primitives with zero mean. Set $h_0=f$, and recursively defineEvery $h_r$ has zero mean. Part (i) gives $h_{r+1}^{\prime }=-h_r$ almost everywhere, and $h_{r+1}(0)=h_{r+1}(1)$ because $R{h}_r$ vanishes at both endpoints. Its periodic extension is consequently locally an absolutely continuous function, with no jump at the integers. Part (ii) gives the uniform estimateThe weak derivative identity on the whole line is $D{\overline{h}}_{r+1}=-{\overline{h}}_r$. After scaling by the integer $j$ and repeating integration by parts $N$ times,The boundary terms vanish because $\varphi$ has compact support in $(0,1)$. The interval contains exactly $j$ periods, soTaking $N=m+1$ therefore yieldsThis rapid distributional decay of periodic oscillations uses only integrability of $f$ and smoothness of the test function.

Let $M={\sup }_j{j}^{-m}\Vert g_j{\Vert }_{L^1}<\infty$. For each $j$, apply the periodic primitives with zero mean construction of part (iii) to $g_j$, writing $h_{j,0}=g_j$ andThe estimates are independent of $j$: $\Vert h_{j,r}{\Vert }_{L^1}\le \Vert g_j{\Vert }_{L^1}$. With $N=m+1$, the same repeated integration by parts gives, for every test function $\varphi \in \mathcal{D}(0,1)$,This is precisely weak convergence of distributions, so $\boxed{u_j\to 0\text{ in }{\mathcal{D}}^{\prime }(0,1)}$. The extra integration by parts beyond the growth exponent $m$ is what ensures decay.