Consider the Heaviside step function: $$H:\mathbb{R}\to \mathbb{R} $$ defined by $$H(x)=\begin{cases} 0 & \mbox{if } x<0 \\ 1 & \mbox{if } x\geq 0\end{cases}$$

Fix any $\delta>0$. Given any $\epsilon >0$, does there exist a real analytic function $F$ such that \begin{align} \sup_{\mathbb{R}}|F(x)-H(x)|<\epsilon\quad\quad \ldots (1) \end{align} and the complexification $\tilde{F}$ of $F$ has no singularities in $\{z\in\mathbb{C}:Im(z)<\delta\}$?

Note: One can use functions like $$F(x)=\frac{1}{1+e^{-kx}}$$ This should satisfy the proximity condition (1) for a sufficiently large $k$ but the singularity condition in the complex domain fails (Has a singularity at $z=\frac{\pi i}{k}$).

  • $\begingroup$ $x / \sqrt{1+x^2}$ approximates $x/|x|,$ then add 1 and divide by 2 $\endgroup$ – Will Jagy Sep 2 '14 at 0:50
  • $\begingroup$ $\frac{kx}{\sqrt{1+k^2x^2}}$ approximates $\frac{x}{|x|}$ but doesn't $\frac{kz}{\sqrt{1+k^2z^2}}$ have a singularity at $\frac{i}{k}$? $\endgroup$ – user42388 Sep 2 '14 at 1:00
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    $\begingroup$ From a distributional sense, one can think of the Step function as the integral of the Dirac Delta function; $H(x) = \int_{-\infty}^x\delta(t)dt$. So, if you think of the delta function as the limiting case of a Gaussian function as the spread goes to $0$, then you might be able to show that $H(x)$ is the limiting case of the Gaussian integral, which is analytic. $\endgroup$ – MrSlunk Sep 2 '14 at 1:06
  • $\begingroup$ $e^{-e^{-kx}}$ seems to be working for a very large $k$ since $e^{-e^{-kz}}$ is entire(?). Can some one confirm? $\endgroup$ – user42388 Sep 2 '14 at 1:17
  • $\begingroup$ But of course (1) fails when $\epsilon < 1/2$. $\endgroup$ – GEdgar Aug 22 '15 at 15:32

The answer is obviously no, even if you relax $F$ to be continuous.

Your requirement (1) asks for a sequence of continuous functions that approximate $H$ uniformly on $\mathbb{R}$. That is impossible since $H$ is not continuous.

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