Past exam of the distribution theory course of the University of Oxford / 2023 / 1

A distribution on $(a,b)$ is a continuous linear functional on the space of test functions $\mathcal{D}(a,b)=C_c^{\infty }(a,b)$. It has order of a distribution at most $m$ if, for every compact set $K\subset (a,b)$, there is $C_K$ such thatThe order of a distribution is the least such $m$. When saying “of order $m$” means merely that this estimate holds, the wording allows a smaller least order.

For each such $K$, the cutoff function theorem supplies a nonnegative $\chi \in \mathcal{D}(a,b)$ equal to one near $K$. If $M=\Vert \varphi {\Vert }_{\infty }$ and $\mathrm{supp}\varphi \subseteq K$, then $M\chi \pm \varphi \ge 0$ everywhere. Positivity of the linear functional $V$ therefore givesHere $V(\chi )\ge 0$ is finite and independent of $\varphi$. This is precisely the local bound for an order-zero distribution, so the given positive distribution is continuous on the space of test functions. The zero distribution also satisfies the estimate.

The distributional derivative of $u$ is defined byIt is a distribution because $\varphi \mapsto {\varphi }^{\prime }$ is a continuous map of the space of test functions into itself. If $f\in C^1(a,b)$ represents a regular distribution, integration by parts givesThe boundary term vanishes because $\varphi$ has compact support. Thus the distributional derivative agrees with the ordinary derivative.

For $\mathrm{supp}\varphi \subseteq K$, the assumed order of a distribution estimate givesConsequently $\boxed{\mathrm{ord}(u^{\prime })\le m+1}$.

On each interval $[k,k+1]$, $k\in \mathbb{Z}$, the function is $d(x)=x-k$ on $[k,k+\frac{1}{2}]$ and $d(x)=k+1-x$ on $[k+\frac{1}{2},k+1]$. Its ordinary derivative exists off $\frac{1}{2}\mathbb{Z}$ and is the locally integrable functionAssign arbitrary values at the endpoints. For a test function $\varphi$, split $-\int d{\varphi }^{\prime }$ at these corners and apply integration by parts on each piece. The adjacent boundary terms cancel because $d$ is continuous, and only finitely many pieces meet the compact support of $\varphi$. HenceThe distributional derivative is exactly the regular distribution of the ordinary derivative defined almost everywhere. There are no Dirac delta distributions in this first distributional derivative: corners change slope, while a jump of the function itself would create a point mass.

The bound $0\le d_n\le 2^{-n-1}$ proves uniform convergence of the series by the Weierstrass M-test. Each summand is continuous and satisfies $d_n(x+1)=d_n(x)$, so its sum $t$, the Takagi function, is continuous and has period one.

For $n\ge m$, both $2^n{u}_m$ and $2^n{v}_m$ are integers, giving $d_n(u_m)=d_n(v_m)=0$. ThereforeThe corners of $d_n$ occur on the grid $2^{-n-1}\mathbb{Z}$. When $n<m$, that grid is contained in $2^{-m}\mathbb{Z}$, so the interior of $[u_m,v_m]$ contains no corner of $d_n$. Its slope on this interval is consequently $+1$ or $-1$.

The interval $[u_{m+1},v_{m+1}]$ is a half of $[u_m,v_m]$. All the old slopes persist on this half, and the new summand contributes one more slope in $\{\pm 1\}$. ThusIf $t$ had a finite derivative $L$ at $x_0$, write $t(y)=t(x_0)+L(y-x_0)+r(y)$ with $r(y)=o(|y-x_0|)$. Since $u_m\le x_0<v_m$ and both endpoints tend to $x_0$,Indeed $|v_m-x_0|+|u_m-x_0|=v_m-u_m$; when $u_m=x_0$, its remainder is zero. This is the secants straddling a differentiability point argument, including dyadic $x_0$. But the displayed increments prevent $S_m$ from converging. Hence $\boxed{t\text{ is nowhere differentiable}}$, using periodicity to cover all real $x_0$.

For the distributional derivative, set $r_n(x)=s(2^{n}x)$, using $s$ from part (iii). The change of variable in the distributional derivative pairing gives $d_n^{\prime }=r_n$; these are the Rademacher functions. Uniform convergence implies weak convergence of distributions, and the continuity of distributional differentiation allows termwise differentiation in that topology. Explicitly,This series assertion is a distributional convergence statement. Its remainder satisfies the concrete estimateIt supplies a well-defined distributional derivative even though the ordinary derivative exists nowhere.

Choose $R>0$ with $\mathrm{supp}\varphi \subset [-R,R]$. Evenness of $f$ changes the symmetrically truncated integral intoThe mean value theorem gives $|\varphi (x)-\varphi (-x)|\le 2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Since $f$ is continuous, it is bounded on $[0,R]$. Thus the Cauchy principal value exists andThis expression is linear in the test function and satisfiesFor each fixed compact support, choose one $R$ once and for all. The estimate proves that the even-weight principal-value distribution has order of a distribution at most one.

If $\varphi (0)=0$, the mean value theorem also gives $|\varphi (x)|\le |x|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Hence $f(x)\varphi (x)/x$ is absolutely integrable near zero and has compact support. The truncated integrals tend to its ordinary Lebesgue integral, so

No. Consider the continuous even functionThe formulas agree at $|x|=e^{-1}$, and $f(x)\to 0$ at zero. Part (i) defines its even-weight principal-value distribution. However,If this were a regular distribution represented by $g\in L_{\mathrm{loc}}^1(\mathbb{R})$, testing on either side of zero would force $g=f/x$ almost everywhere there, by uniqueness of the locally integrable function representing a regular distribution. Changing the value at zero cannot remove the displayed divergence, contradicting local integrability of $g$.

More generally, the exact condition isNecessity is the preceding argument; sufficiency follows because $f/x$ is then a locally integrable function and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.