The Heaviside Step Function is given by:
$$ \begin{align*} H(x) = \Theta = \left\{\begin{array}{c} 0,if \quad x<0 \\ 1, if \quad x \geq 0 \end{array}\right. \end{align*} $$
I should now proof that the absolute condition number is $\infty$. I know there is this equation:
$‖f (\tilde{x}) − f (x)‖ ≤ k_{abs} ‖\tilde{x} − x‖ + σ(‖\tilde{x} − x‖)$
and I normally know how to calculate the absolute condition number but since this function is not continuously differentiable I have no clue how to start with this task. Thanks for your help.
If both $x$ and $\tilde x$ are $\ge 0$, you have that $f(x)=f(\tilde x)=1$ and so,
$$ |f(x)-f(\tilde x)| = 0 \leq k|x-\tilde x| $$
holds for any $k$. The same happens in $x,\tilde x <0$.The actual restrictions on $k$ appear when $x$ and $\tilde x$ have different signs. In that case, the inequality $$ |f(x)-f(\tilde x)| = 1 \leq k |x-\tilde x| $$
can only hold for all such $x, \tilde x$ if $k = \infty$. For instance,
$$ |f(-1/n) - f(1/n)| = 1 \leq \frac{2k}{n} $$
implies that we should have $k\ge \frac n2$ for all $n$.