Fundamental Solution to Nth Derivative

Theorem
Let $H$ be the Heaviside step function.

Let $n \in \N_{\mathop > 0}$.

Let $\ds \map {f_n} x = \map H x \frac {x^{n - 1}} {\paren {n - 1}!}$.

Let $T_{f_n}$ be the distribution associated with $f_n$.

Then, in the distributional sense, $T_{f_n}$ is the fundamental solution of


 * $\dfrac {\rd^n} {\rd x^n} T_{f_n} = \delta$

Proof
Proof by Principle of Mathematical Induction:

Basis for the Induction
Let $n = 1$.

By Distributional Derivative of Heaviside Step Function:


 * $\dfrac \rd {\rd x} T_{f_1} = \delta$

Induction Hypothesis
This is our induction hypothesis:


 * $\dfrac {\rd^k} {\rd x^k} T_{f_k} = \delta$

Now we need to show true for $n = k + 1$:


 * $\dfrac {\rd^{k + 1}} {\rd x^{k + 1}} T_{f_{k + 1}} = \delta$

Induction Step
This is our induction step:

$\ds x \stackrel f {\longrightarrow} \map H x \frac {x^k} {k!}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.

By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.

Moreover:


 * $\ds x < 0 \implies \paren {\map H x \frac {x^k} {k!}}' = 0$.


 * $\ds x > 0 \implies \paren {\map H x \frac {x^k} {k!}}' = \frac {x^{k - 1}}{\paren {k - 1}!}$.

Altogether:


 * $\ds \forall x \in \R \setminus \set 0 : \paren {\map H x \frac {x^k} {k!}}' = \map H x \frac {x^{k - 1}}{\paren {k - 1}!}$

Furthermore:


 * $\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \frac {\paren {0^+}^k} {k!} - \map H {0^-} \frac {\paren {0^-}^k} {k!} = 0$

By the Jump Rule:


 * $T'_{\map H x \frac {x^k} {k!}} = T_{\map H x \frac {x^{k - 1}}{\paren {k - 1}!}}$

Suppose:


 * $\dfrac {\rd^n} {\rd x^k} T_{f_k} = \delta$

The result follows by induction.