Past exam of the distribution theory course of the University of Oxford / 2023 / 3 / b / iii / Solution

Because $f$ is a compactly supported distribution, the convolution of distributions with a compactly supported factor defines $E\ast f$. The convolution derivative identity and part (i) giveThe difference between any two solutions has zero ${\partial }_{\bar{z}}$ distributional derivative. Part (ii) therefore proves that the general solution of the partial differential equation isThis formula is an identity of distributions; a pointwise integral is not required for arbitrary $f$.

If $u$ were a smooth function, its Wirtinger derivative ${\partial }_{\bar{z}}u=f$ would also be smooth, contradicting $\mathrm{singsupp}f\ne \varnothing$. In fact the singular support is exactly preserved here:One inclusion follows because a differential operator cannot create a singularity where $u$ is smooth. For the reverse inclusion, $E$ is smooth off zero, and its singular support is exactly $\{0\}$: if it were smooth near zero, ${\partial }_{\bar{z}}E={\delta }_0$ would be smooth there. The support and singular support rules for convolution give $\mathrm{singsupp}(E\ast f)\subseteq \{0\}+\mathrm{singsupp}f$, and adding the smooth entire function $h$ leaves this singular support unchanged. Thus there is no globally smooth solution when the given singular support is nonempty.