Question

Let \(f\) and \(g\) be continuously differentiable and such that \(f, g, f^{\prime}\), and \(g^{\prime}\) are absolutely integrable on the interval \([0, \infty)\), and \(f(0)=g(0)=0\). Prove that $$ (f * g)^{\prime}=f^{\prime} * g=f * g^{\prime} . $$

Short answer

We applied the Leibniz rule and properties of convolution to prove \((f * g)' = f' * g = f * g'\).

Step-by-step solution

Understanding Convolution

The convolution of two functions, \(f * g\), is defined as \((f * g)(t) = \int_0^t f(t-x)g(x) \, dx\). It's a way to combine two functions into a single function, showcasing the impact of one signal on another.

Apply Differentiation to Convolution

To prove \((f * g)' = f' * g\), apply the Leibniz rule for differentiating under the integral sign: \(\frac{d}{dt} \int_0^t f(t-x)g(x) \, dx = f(0)g(t) + \int_0^t f'(t-x)g(x) \, dx - f(t)g(0)\). Given \(f(0) = g(0) = 0\), this simplifies to \((f * g)'(t) = \int_0^t f'(t-x)g(x) \, dx\), which is \(f' * g\).

Use Symmetry to Prove Equivalence

To show \(f' * g = f * g'\), we swap roles of \(f\) and \(g\). We know from the commutative property of convolution that \((f * g)(t) = (g * f)(t)\). Thus, \(f' * g = g * f' = f * g'\).

Final Equality Statement

Thus, we have shown \((f * g)' = f' * g\) using Leibniz rule and \(f' * g = f * g'\) using commutative property. Therefore, \((f * g)' = f' * g = f * g'\), completing the proof.

Key concepts

Differentiation under the integral sign

Differentiation under the integral sign is a powerful mathematical technique that allows us to differentiate an integral. This method is crucial when dealing with functions defined by integrals, especially when parameters within the integrals need to be differentiated. This concept is largely attributed to Leibniz, and hence it's sometimes called the Leibniz integral rule. In simple terms, this technique lets you "move the differentiation operator," \(\frac{d}{dt}\), inside the integral:

• Consider an integral of the form \(\int_{a(t)}^{b(t)} f(x, t) \, dx\).
• To differentiate \(\int_{a(t)}^{b(t)} f(x, t) \, dx\) with respect to \(t\), we apply the Leibniz rule: \[\frac{d}{dt} \int_{a(t)}^{b(t)} f(x, t) \, dx = f(b(t), t) \cdot \frac{d}{dt}b(t) - f(a(t), t) \cdot \frac{d}{dt}a(t) + \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t} \, dx\]

In the original exercise, since \(a(t)\) and \(b(t)\) are constants, only the last term remains, simplifying the differentiation. If the integral resembles the convolution structure in the exercise, this rule facilitates finding the derivative of the convolution of two functions by adjusting the order of integration and differentiation.

Commutative property of convolution

The commutative property of convolution is one of the fundamental characteristics of this operation. In mathematics, when something is commutative, you can swap the order of the operands without changing the result. For convolution, it means that \(f * g = g * f\). This property simplifies many proofs and calculations, particularly in signal processing and system analysis.

• This means if you are convolving two functions \(f\) and \(g\), you can do this either as \((f * g)(t)\) or \((g * f)(t)\).
• From the perspective of integrating one function with another, this property arises because integration itself is commutative over the real line.
  • The expression \(\int_{0}^{t} f(t-x)g(x) \, dx\) can be rewritten as \(\int_{0}^{t} g(t-x)f(x) \, dx\) by simply changing variables.

In the context of the given exercise, recognizing that convolution is commutative helps show that \(f' * g = f * g'\). This commutativity imbues the convolution operation with flexibility, allowing us to interchange the functions, which is key to demonstrating the equivalence of different derivative expressions of convoluted functions.

Leibniz rule

The Leibniz rule, named after the mathematician Gottfried Wilhelm Leibniz, is a formula used to differentiate products of functions. While it's classically known for differentiating products, its powerful variant allows us to differentiate integrals.

• In the context of integration, Leibniz rule states that you can differentiate under the integral sign when the boundary limits are constants and/or the integrated function depends on the parameter explicitly.
• This rule takes the form: \[\frac{d}{dt} \int_{a}^{b} f(x, t) \, dx = \int_{a}^{b} \frac{\partial}{\partial t}f(x, t) \, dx\]
• This means you differentiate the integrand function as usual inside the integral.

In the case of the exercise, using Leibniz rule makes it easier to handle and differentiate the convolution integral, which involves integrands that change with respect to a variable. Applying this rule streamlines the process of finding \( (f * g)' = f' * g \), by turning a potentially complicated operation into a manageable computation.