What's the essential bandwidth of the unit step function?
there can be no essential bandwidth of the unit step function.  defining such would lead you to self-contradiction.

suppose you have a function $x(t)$ which has Fourier Transform

$$ X(f) = \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ dt $$

and from $X(f)$, you defined some consistent measure of bandwidth $B_x$ in such a way that is independent of the amplitude of $X$ and is dimensionally consistent with $f$.

now suppose you scale the time argument of $x(t)$ to speed it up or slow it down:

$$ y(t) = x(a t) \quad \text{for } a &gt; 0 $$

we know that the the Fourier Transform of $y(t)$ would be

$$ Y(f) = \frac{1}{a} X\left(\frac{f}{a} \right) $$

and the bandwidth of $Y(f)$ defined in the same manner as $B_x$ would be

$$ B_y = a B_x $$

so, suppose $a&gt;1$, then you speed up $x(t)$ by the factor $a$ and the bandwidth is increased by the same factor.

but if you "speed up" (or slow down) the unit step function, you still have the very same unit step function:

$$ u(t) = u(a t) \quad \text{for } a &gt; 0 $$

where

$$ u(t) \triangleq \begin{cases}
1, &amp; \text{if } t &gt; 0 \\
0, &amp; \text{if } t &lt; 0 \\
\end{cases} $$ 

so scaling time for the unit step does not change it, yet if it had a bandwidth, scaling the time must change that bandwidth.  a contradiction results.