Convolution of Real Function with Rectangle Function

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Theorem

Let $f: \R \to \R$ be a bounded piecewise continuous real function.

Consider the rectangle function $\Pi: \R \to \R$.

Then:

$\forall x \in \R: \map \Pi x * \map f x = \ds \int_{x \mathop - \frac 1 2}^{x \mathop + \frac 1 2} \map f u \rd u$

where $*$ denotes the convolution integral.


Proof

By definition of convolution integral:

$\ds \map \Pi x * \map f x = \int_{-\infty}^\infty \map \Pi {x - t} \map f x \rd t$

The above integral exists because $f$ is discontinuous on a countable set (see Lebesgue-Vitali Theorem).

This is equal to:

$\ds \int_{-\infty}^{x - \frac 1 2} \map \Pi {x - t} \map f t \rd t + \int_{x - \frac 1 2}^{x + \frac 1 2} \map \Pi {x - t} \map f t \rd t + \int_{x + \frac 1 2}^\infty \map \Pi {x - t} \map f t \rd t$

By definition of the rectangle function we have:

$\map \Pi {x - t} = 1 \iff \size {x - t} \le \dfrac 1 2$
$\map \Pi {x - t} = 0 \iff \size {x - t} > \dfrac 1 2$

Therefore:

\(\ds \int_{-\infty}^{x - \frac 1 2} \map \Pi {x - t} \map f t \rd t\) \(=\) \(\ds \int_{x + \frac 1 2}^\infty \map \Pi {x - t} \map f t \rd t\)
\(\ds \) \(=\) \(\ds 0\)

What remains is:

\(\ds \int_{x - \frac 1 2}^{x + \frac 1 2} \map \Pi {x - t}\map f t \rd t\) \(=\) \(\ds \int_{x - \frac 1 2}^{x + \frac 1 2} \map f t \rd t\)
\(\ds \leadsto \ \ \) \(\ds \map \Pi x * \map f x\) \(=\) \(\ds \int_{x - \frac 1 2}^{x + \frac 1 2} \map f t \rd t\)

$\blacksquare$


Sources