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

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.