Question

Let \(f, g\) be defined on \(\mathbb{R}^{+}\)in such a way that \(f\) and \(g\) are Riemann-integrable on each interval \([0, s], s>0 .\) Set \(F(s)=\) \(\int_{0}^{s} f(s) d s\) and \(G(s)=\int_{0}^{s} g(s) d s, s \in \mathbb{R}^{+} .\)Define the convolution \(f * g \mathrm{by}\) $$ (f * g)(s)=\int_{0}^{s} f(s-t) g(t) d t, \quad s \geq 0 $$ By assuming that the theorem on repeated intergrals is applicable, prove that $$ \int_{0}^{a}(f * g)=(F * g)(a), \quad a \geq 0 $$ and also that \((f * g) * h=f *(g * h)\)

Short answer

In summary, we have proven that the convolution operation has the following two properties: 1. \(\int_{0}^{a}(f * g) = (F * g)(a)\), where the capital \(F\) indicates the integral of \(f\), and \(s \geq 0\). 2. Convolution is associative: \((f * g) * h = f * (g * h)\). The proof was carried out in five steps that involved finding the expressions for the convolutions, verifying their equality and using Fubini's theorem to change the order of integration, ultimately demonstrating that both expressions were identical.

Step-by-step solution

Find the expression for \((F * g)(a)\).

We are already given the expression for convolution: $$ (f * g)(s) = \int_{0}^{s} f(s-t) g(t) dt, \quad s \geq 0 $$ Now, we need to find the expression for \((F * g)(a)\). We can obtain this by substituting \(F(s) = \int_{0}^{s} f(s) ds\). $$ (F * g)(a) = \int_{0}^{a} F(a-t) g(t) dt $$

Substitute \((f * g)(a)\) and \((F * g)(a)\) into the expression and show they are equal.

We want to show that \(\int_{0}^{a}(f * g) = (F * g)(a)\). Let us rewrite the left side of the equation: $$ \int_{0}^{a}(f * g) = \int_{0}^{a} \int_{0}^{s} f(s-t) g(t) dt ds $$ Now we use Fubini's theorem, which allows us to switch the order of integration: $$ \int_{0}^{a} \int_{0}^{s} f(s-t) g(t) dt ds = \int_{0}^{a} \int_{0}^{a-t} f(s-t) g(t) ds dt $$ If we perform the substitution taking \(u = s - t\), we get \(s = u + t\) and \(ds = du\). Thus, the integral becomes: $$ \int_{0}^{a} \int_{0}^{a-t} f(s-t) g(t) ds dt = \int_{0}^{a} \int_{t}^{a} f(u) g(t) du dt = \int_{0}^{a} F(a - t) g(t) dt $$ Now we can see that the left side of the original equation equals to the right side, \(\int_{0}^{a}(f * g) = (F * g)(a)\).

Find the expression for \((f * g) * h\).

We can calculate \((f * g) * h\) by following the convolution formula: $$ (f * g) * h = \int_{0}^{s} (f * g)(s-t) h(t) dt = \int_{0}^{s} \left(\int_{0}^{s-t} f(u) g(s-t-u) du\right) h(t) dt $$

Find the expression for \(f * (g * h)\).

Similarly, we can find the expression for \(f * (g * h)\): $$ f * (g * h) = \int_{0}^{s} f(s-t)\left(\int_{0}^{t} g(t-u) h(u) du\right) dt $$

Show that both expressions for points 3 and 4 are equal.

We have the two expressions for convolutions, and we need to show that they are equal: $$ (f * g) * h = \int_{0}^{s} \left(\int_{0}^{s-t} f(u) g(s-t-u) du\right) h(t) dt $$ and $$ f * (g * h) = \int_{0}^{s} f(s-t)\left(\int_{0}^{t} g(t-u) h(u) du\right) dt $$ Using Fubini's theorem, we can change the order of integrals for both expressions: For \((f * g) * h\): $$ \int_{0}^{s} \left(\int_{0}^{s-t} f(u) g(s-t-u) du\right) h(t) dt = \int_{0}^{s}\int_{0}^{s-u} f(u) g(v) h(s-u-v) dv du $$ For \(f * (g * h)\): $$ \int_{0}^{s} f(s-t)\left(\int_{0}^{t} g(t-u) h(u) du\right) dt = \int_{0}^{s}\int_{0}^{s-u} f(u) g(v) h(s-u-v) dv du $$ We can see that both expressions are equal. Therefore, the convolution operation is associative: \((f * g) * h = f * (g * h)\).

Key concepts

Riemann Integrable

Understanding when a function is Riemann integrable is fundamental in calculus, especially when we're working with convolutions or any form of integration. A function is considered to be Riemann integrable if you can define its integral over an interval as the limit of sums of function values times increments of partitions, as the maximum partition size approaches zero. In simpler terms, if you can slice the region under the curve into smaller and smaller rectangles and their total area approaches a single value, the function is Riemann integrable.

This property is essential because it assures that the area under the curve has a definite value. In our problem, both functions, \(f\) and \(g\), are assumed Riemann integrable, which means we can proceed with various integration techniques to find the convolution of the functions without worrying about the existence of the integrals themselves.

Fubini's Theorem

Fubini's theorem is a cornerstone of multiple integration that comes into play when evaluating iterated integrals of functions in more than one variable. Essentially, the theorem states that if a function is integrable over a rectangle, the order of integration can be switched without affecting the result. For the theorem to apply, the function has to be integrable over the set (such as a rectangle in two dimensions) and usually either non-negative or absolutely integrable over the set.

In the context of our exercise, Fubini's theorem allows us to change the order in which we perform the integration when finding the convolution, \((f * g)(s)\), of two functions. This is critical to simplify the problem and ultimately to show that the convolution is associative — a property we want to prove.

Repeated Integrals

Repeated integrals refer to the situation where we're computing the integral of an integral; essentially, integrating a function multiple times over different variables or over the same variable in succession. In practice, this means performing an integration step, then taking that result and integrating it again.

In the provided exercise, repeated integration comes into play because convolution involves an integral of an integral, as seen in the definition of \((f * g)(s)\). The outer integral integrates over the result of the inner integral, which itself is the product of two functions within an integration. Understanding how to compute repeated integrals correctly, and the conditions under which they can be computed (like those provided by Fubini's theorem), is key to solving problems involving convolutions.

Associative Property

The associative property is a basic principle of mathematics that refers to the idea that when you have a sequence of operations of the same kind, the way you group these operations does not change the result. In arithmetic, this is seen with addition and multiplication: for instance, \((a + b) + c = a + (b + c)\).

In the context of convolutions, the associative property suggests that the order in which you convolute multiple functions does not change the outcome. This implies that \((f * g) * h\) will yield the same result as \(f * (g * h)\), given that \(f\), \(g\), and \(h\) are functions that satisfy the conditions required for convolution. Proving this property as part of our exercise demonstrates that convolutions behave predictably under grouping, which is hugely significant in both theoretical and applied mathematics.