Continuity of Heaviside Step Function

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Theorem

Let $\mu_c: \R \to \R$ be the Heaviside step function:

$\map {\mu_c} x = \begin {cases}

0 & : x < c \\ 1 & : x > c \\ \text {arbitrary} & : x = c \end {cases}$

Then $\mu_c$ is continuous at every point of $\R$ except at $c$.


Proof

Let $x \in \R: x \ne c$.

Let $\epsilon \in \R_{>0}$.

Let $\delta < \size {x - c}$.

Then by definition of the Heaviside step function:

$\forall y \in \closedint {x - \delta} {x + \delta}: \map {\mu_c} x = \begin {cases}

0 & : x < c \\ 1 & : x > c \end {cases}$

Thus:

$\forall y \in \closedint {x - \delta} {x + \delta}: \map {\mu_c} y = \map {\mu_c} x$

Thus:

$\size {y - x} \implies: \map {\mu_c} y - \map {\mu_c} x < \epsilon$

and so $\mu_c$ is continuous at $\R$ except at $x$.

$\Box$


Now suppose $x = c$.

Let $\epsilon = \dfrac 1 3$.

Let $\delta \in \R_{>0}$.

Let $\alpha \in \R_{>0}: \alpha < \delta$.

Let $y = x + \alpha$.


Let $y \in \R: \size {y - x} < \delta$ and $y \ne x$.


Suppose $\map {\mu_c} y - \map {\mu_c} x < \epsilon$.


Then:

\(\ds y\) \(=\) \(\ds x + \alpha\)
\(\ds \leadsto \ \ \) \(\ds \map {\mu_c} y\) \(=\) \(\ds 1\)


while:

\(\ds y\) \(=\) \(\ds x - \alpha\)
\(\ds \leadsto \ \ \) \(\ds \map {\mu_c} y\) \(=\) \(\ds 0\)


So whatever the value of $\map f x$, either:

$\map f {x + \alpha} - \map f x > \epsilon$

or:

$\map f {x - \alpha} - \map f x > \epsilon$

It follows that $\mu_c$ is not continuous at $c$.

$\blacksquare$


Also see


Sources

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