Step Function satisfies Dirichlet Conditions

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Theorem

Let $\alpha, \beta \in \R$ be a real numbers such that $\alpha < \beta$.

Let $\map f x$ be a step function defined on the interval $\openint \alpha \beta$.


Then $f$ satisfies the Dirichlet conditions.


Proof

Recall the definition of step function:

A real function $f: \R \to \R$ is a step function if and only if it can be expressed as a finite linear combination of the form:

$\map f x = \lambda_1 \chi_{\mathbb I_1} + \lambda_2 \chi_{\mathbb I_2} + \cdots + \lambda_n \chi_{\mathbb I_n}$

where:

$\lambda_1, \lambda_2, \ldots, \lambda_n$ are real constants
$\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ are open intervals, where these intervals partition $\R$ (except for the endpoints)
$\chi_{\mathbb I_1}, \chi_{\mathbb I_2}, \ldots, \chi_{\mathbb I_n}$ are characteristic functions of $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$.


Recall the Dirichlet conditions:

\((\text D 1)\)   $:$   $f$ is absolutely integrable      
\((\text D 2)\)   $:$   $f$ has a finite number of local maxima and local minima      
\((\text D 3)\)   $:$   $f$ has a finite number of discontinuities, all of them finite      


We inspect the Dirichlet conditions in turn.


$(\text D 1)$: Absolute Integrability

From Definite Integral of Step Function:

$\ds \int_\alpha^\beta \map f x \rd x = \sum_{k \mathop = 1}^n \lambda_k \paren {\beta_k - \alpha_k}$

where $\alpha_k, \beta_k$ are the endpoints of $\mathbb I_k$ for $1 \le k \le n$.

When $\lambda_k < 1$, we have that:

$\size {\lambda_k} = -\lambda_k$

and so:

$\ds \int_{\alpha_k}^{\beta_k} \lambda_k \rd x = -\lambda_k \paren {\beta_k - \alpha_k}$

Absolute integrability follows.

$\Box$


$(\text D 2)$: Finite Number of Local Maxima and Minima

By nature of $f$ being a step function, the image set of $f$ is finite:

$f \openint \alpha \beta = \set {\lambda_1, \lambda_2, \ldots, \lambda n}$

The set of local maxima and the set of local minima must be subsets of $f \openint \alpha \beta$.

Therefore there must be a finite number of each.

$\Box$


$(\text D 3)$: Discontinuities are Finite in Number and Nature

From Constant Real Function is Continuous, there are no discontinuities except perhaps at the endpoints of $\mathbb I_k$ for $1 \le k \le n$.

Let us inspect those endpoints.

Let $\alpha_k, \beta_k$ be the endpoints of $\mathbb I_k$ for $1 \le k \le n$, where $\alpha_k < \beta_k$.

Let $\mathbb I_1, \mathbb I_2, \ldots, \mathbb I_n$ be ordered such that:

$\forall k: \alpha_k < \alpha_{k + 1}$

By definition, $\beta_{k - 1} = \alpha_k$.

Thus we have that:

\(\ds \lim_{x \mathop \to {\beta_{k - 1} }^-}\) \(=\) \(\ds \lim_{x \mathop \to {\alpha_k}^-}\)
\(\ds \) \(=\) \(\ds \lambda_{k - 1}\)
\(\ds \lim_{x \mathop \to {\alpha_k}^+}\) \(=\) \(\ds \lambda_k\)

and if $\lambda_k \ne \lambda_{k - 1}$ then this is indeed a finite discontinuity.

By the nature of the step function, there are a finite number of them.

$\Box$


Thus all Dirichlet conditions are satisfied by $f$.

$\blacksquare$


Sources

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