Mollification of a complex two-step function

The input is bounded and supported in $[0,2]$, so F3 makes it integrable and locally integrable. In the convolution integral, $f(x-t)$ has real part one exactly for $t\in [x-1,x]$ and imaginary part one exactly for $t\in [x-2,x-1]$. Integrating the two components gives the displayed formula. Their common endpoint has measure zero by F3. F1 makes the output smooth. Since ${\rho }_{\epsilon }$ vanishes outside $[-\epsilon ,\epsilon ]$, both integrals vanish for $x<-\epsilon$ and for $x>2+\epsilon$, proving the stated support inclusion.

For $0<\epsilon <1\mathrm{/}4$, put $U={\bigcup }_{a\in \{0,1,2\}}[a-\epsilon ,a+\epsilon ]$. Outside $U$, the entire interval $[x-\epsilon ,x+\epsilon ]$ stays in a constant region of $f$, so the unit kernel mass from F1 implies $({\rho }_{\epsilon }\ast f)(x)=f(x)$. Each component of the convolution is between zero and one, because ${\rho }_{\epsilon }\ge 0$ and its integral is one. The same is true for each component of $f$, including its value at the shared endpoint. Consequently the modulus error is at most $\sqrt{2}\le 2$ everywhere and is zero outside $U$. F3–F4 give $\mu (U)\le 6\epsilon$. Since $\mathrm{\mid }{\rho }_{\epsilon }\ast f-f{\mathrm{\mid }}^p\le 2^p{\mathbf{1}}_U$, F5–F6 and the definition of $N_p$ in F2 give $\mathrm{\parallel }{\rho }_{\epsilon }\ast f-f{\mathrm{\parallel }}_p^p\le 2^p\mu (U)\le 2^{p}6\epsilon$. Taking p-th roots proves the bound, and its limit is zero for each fixed finite $p$.

Let $h$ be any continuous complex function. If $\mathrm{\parallel }h-f{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$, choose a real essential bound $a<1\mathrm{/}2$ for the error. On $(-1\mathrm{/}2,0)$ the function $f$ is zero, so $\mathrm{\mid }h(x)\mathrm{\mid }\le a$ a.e.; by continuity this inequality holds throughout that interval, since any failure persists on an open interval of positive measure by F3. Similarly $\mathrm{\mid }h(x)-1\mathrm{\mid }\le a$ throughout $(0,1\mathrm{/}2)$. Taking the respective limits at zero gives $\mathrm{\mid }h(0)\mathrm{\mid }\le a$ and $\mathrm{\mid }h(0)-1\mathrm{\mid }\le a$, whence $1\le \mathrm{\mid }h(0)\mathrm{\mid }+\mathrm{\mid }h(0)-1\mathrm{\mid }\le 2a<1$, a contradiction. Thus every such $h$ has error at least $1\mathrm{/}2$, in particular the smooth function from step 1.1.