Distributional Derivative of Floor Function

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Theorem

Let $\floor x$ be the floor function.

Let $\map {\operatorname {III}} x$ be the Dirac comb.


Then the distributional derivative of $\floor x$ is $\map {\operatorname {III}} 0$.


Proof

By definition:

$\floor x := \sup \set {m \in \Z: m \le x}$

Hence, $\forall m \in \Z : \forall x \in \openint m {m + 1}$ the floor function is constant.

Therefore:

$\forall m \in \Z : \forall x \in \openint m {m + 1} : \dfrac {\d \floor x} {\d x} = 0$

Every $x \in \Z$ is a discontinuity of $\floor x$.

Hence, the jump rule has to be applied to each such $x$.

Suppose $k \in \Z$.

Then:

$\ds \lim_{x \to k^+} \floor x = k$
$\ds \lim_{x \to k^-} \floor x = k - 1$

By the Jump Rule:

\(\ds T'_{\floor x}\) \(=\) \(\ds \sum_{k \mathop \in \Z} \delta_k\)
\(\ds \) \(=\) \(\ds \map {\operatorname {III} } 0\) Definition of Dirac Comb

$\blacksquare$


Sources