Why is the Heaviside function 1/2 at the origin? When looking at the definition of the Heaviside function it states:
$H(t) = \begin{cases} 0 &\text{if } t<0\\\frac{1}{2} &\text{if }t=0\\1 &\text{if } t>0. \end{cases}$
Where does the $\frac{1}{2}$ come from for $t=0$?
Thanks in advance!
 A: It is merely a convention. In my experience, it is more often defined as $$H(x)=\begin{cases}1,&x\ge 1\\0,&x<0\end{cases}$$or$$H(x)=\begin{cases}1,&x> 1\\0,&x\le0.\end{cases}$$
However, one can try to define this function as pointwise limit (as $n\to\infty$) of the family of continuous functions
$$H_n(x)=\begin{cases}1,&x> \frac 1n\\0,&x<-\frac 1n\\\frac{nx+1}{2},&|x|\le \frac 1n.\end{cases}$$
This definition leads to $H(0)=\lim_{n\to\infty} H_n(0)=\frac 12$.
A: As @TZakrevskiy noted, it's convention. In my experience, it's done for two reasons:


*

*Because Fourier series, and other othogonal function representations of functions, tend to converge in the mean, that is, they converge to:
$$ \overline{f}(x) = \lim_{h\rightarrow 0} \frac{\int_{x-h/2}^{x+h/2} f(x) \operatorname{d}x}{h}.$$
Regardless of the initial convention for the step function, a representation that converges in the mean will produce $H(0) = \frac{1}{2}$.

*Because the useful definition of the step function will usually depend on which definition of the Dirac delta function applies to the problem. For instance, if the appropriate delta function is a Gaussian,
$$ \delta(x)  = \overline{\lim_{\sigma \rightarrow 0}} \frac{1}{\sigma \sqrt{2\pi}} \operatorname{e}^{-\frac{x^2}{2\sigma^2}},$$
then the appropriate step function is:
$$ H(x) = \lim_{\sigma \rightarrow 0} \frac{1 + \operatorname{erf}\left(\frac{x}{\sigma \sqrt{2}}\right)}{2}. $$
I don't know of an example of a delta function that would produce value for $H(0)$ that isn't $\frac{1}{2}$ off the top of my head, but it could be possible.


(I use $\overline{\lim}$ to emphasize that the limit is really only meaningful if it's done outside of an integral that the argument of the limit is inside of).
