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

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 $\boxed{u=1\text{ gives no distributional limit}}$.