8. Convolutions
≪ 7. Comments on Homework 3 | Table of Contents | 9. Multivariable Calculus Refresher ≫A few weeks ago, we saw a proof of the Stone-Weierstrass theorem that used convolutions to prove that polynomials are dense in the space of continuous functions on a compact interval.
In case you are not a convolution-pilled integral-maxxer (which means my indoctrination is failing), we’ll be building a little bit more background on what convolutions are and providing another example of why they’re useful. Now, we are equipped with an additional tool: the space of functions \(C _{0}^{\infty}(\mathbb{R})\), the space of compactly supported smooth functions on the real line. This is perhaps the nicest possible space of functions one could ever hope for, and we’ll be studying how continuous functions are affected by convolutions with such functions.
Compactly Supported Smooth Functions
Let’s refresh our memory of a few definitions.
Definition 1.
Let \(X\) be a metric space and let \(f:X\to \mathbb{R}\) be a function. The support of \(f\), denoted \(\operatorname{supp}f\), is the closure of the set \(\left\lbrace x\in X : f(x)\neq 0 \right\rbrace\). We say \(f\) is compactly supported if \(\operatorname{supp}f\) is compact.
We denote by \(C_0(X)\) the space of compactly supported continuous functions \(X\to \mathbb{R}\). We denote by \(C _{0}^{\infty}\left( \mathbb{R} \right)\) the set of compactly supported smooth functions on the real line.
As you hopefully know, continuous functions on compact metric spaces enjoy some nice properties, most notably that they are uniformly continuous. Unfortunately, we don’t always have the luxury of working with a compact domain all the time; in practise, many domains we run into are open, unbounded, or worse subsets of (\mathbb{R}^n).
Compactly supported (smooth) functions give us a nice bridge between the extremely nice properties of functions on compact spaces and the unpleasant realities of life. For instance,
Proposition 2.
Let \(f\in C_0\left( \mathbb{R}^n \right)\). Then \(f\) is uniformly continuous.
Proof
Let \(f\in C_0\left( \mathbb{R}^n \right)\), and let \(S = \operatorname{supp} f\). Note \(S\) is closed and bounded by Heine-Borel, so in particular there exists some \(R>0\) such that \(S\subseteq \overline{B}_R\), where \[\overline{B}_R = \left\lbrace x\in \mathbb{R}^n : \left\lVert x \right\rVert \leq R \right\rbrace.\] Let \(\epsilon > 0\). Since \(f\) is uniformly continuous on \(\overline{B} _{R+1}\), there exists a \(\delta > 0\) such that for any \(x, y\in \overline{B} _{R+1}\), \(\left\lVert x-y \right\rVert < \delta \) implies \(\left\lvert f(x)-f(y) \right\rvert< \epsilon \).
Consider \(\delta ’ = \min \left( \delta , 1 \right)\), and suppose \(x, y\in \mathbb{R}^n\) satisfy \(\left\lVert x-y \right\rVert < 1\). There are two (possibly overlapping) cases:
- \(x, y\in \overline{B} _{R+1}\), in which case \(\left\lvert f(x)-f(y) \right\rvert < \epsilon \) by definition of \(\delta \).
- \( x, y\notin \overline{B}_R\supseteq \operatorname{supp}f \), in which case \(f(x)=f(y)=0\). Hence, \(\left\lvert f(x)-f(y) \right\rvert = 0 < \epsilon \). These cases are exhaustive: the only case where neither condition is satisfied is when \(x\in \overline{B}_{R}\) and \(y\notin \overline{B} _{R+1}\) (or vice versa). However, this forces \(\left\lVert x-y \right\rVert \geq 1 > \delta ‘\).
Hence, \(\left\lVert x-y \right\rVert < \delta ‘\) implies \(\left\lvert f(x) - f(y) \right\rvert < \epsilon \). \(\square\)
Draw a picture to see what’s going on!
Exercise 3.
Does this generalise to \(f\in C_0(X)\), when \(X\) is an arbitrary metric space? Prove your claim or give a counterexample.
Hint
The last thing we should say is that \(C _{0}^{\infty}\left( \mathbb{R}^n \right)\neq \left\lbrace 0 \right\rbrace\), i.e. there exists a nonzero compactly supported smooth function. This is actually a highly nontrivial fact, and hopefully you guys have covered this fact in class.
It is actually enough to show that \(C _{0}^{\infty} \left( \mathbb{R} \right)\neq \left\lbrace 0 \right\rbrace\), as one can produce a \(\varphi\in C _{0}^{\infty}(\mathbb{R})\) such that \(\varphi\) is constant in a neighbourhood containing \(0\). Then, one can work in polar coordinates on \(\mathbb{R}^n\) and send \(x\mapsto \varphi\left( \left\lVert x \right\rVert \right) \in C _{0}^{\infty}\left( \mathbb{R}^n \right)\). This “radial extension” of \(\varphi\) is a great trick to remember!
Convolutions with Compactly Supported Smooth Functions
We’ll be mostly sticking with the real line for now; multi-dimensional convolutions will share many of the properties that we are about to enumerate, but we have not formally introduced multi-dimensional integrals in this class.
For some notation, if \(\varphi\in C _{0}^{\infty}(\mathbb{R})\), we will write \[\int \varphi(x) dx = \int _{-\infty}^{\infty} \varphi(x) dx = \int _{a}^{b} \varphi(x) dx,\] where \(a, b\) are any real numbers such that \(\operatorname{supp}\varphi\subseteq \left[ a,b \right]\). This will save me a lot of wrist pain.
If \(f\in C(\mathbb{R})\), we recall the definition of the convolution of \(f\) with \(\varphi\): \[f*\varphi(t) = \int f(x) \varphi \left( t-x \right) dx,\] where \(t\) is any real number. As described a while ago, we should think of this as a weighted average of \(f\). Visually, we are “smearing” the graph of \(f\) according to the shape of \(\varphi\) (in fact, this is where “Gaussian blurs” get their name).
Yet another way to describe this is that we are “smoothing out” the graph of \(f\); formally, this smoothness of the convolution will prove to be an invaluable feature.
Proposition 4.
Let \(f\in C(\mathbb{R})\) and \(\varphi \in C _{0}^{\infty}(\mathbb{R})\). Then \(f*\varphi\in C ^{\infty}(\mathbb{R})\).
We’ll leave the proof as an exercise! The smoothness comes from the fact that you can move derivatives under the integral sign. There are several other properties that are good to know as well:
Proposition 5.
Let \(\varphi, \psi \in C _{0}^{\infty}(\mathbb{R})\). Then, \[\operatorname{supp} \left( \varphi* \psi \right) \subseteq \left\lbrace x+y : x\in \operatorname{supp}\varphi,\ y\in \operatorname{supp}\psi \right\rbrace.\]
Proof
Proposition 6.
Let \(f\in C(\mathbb{R})\) and \(\varphi\in C _{0}^{\infty}(\mathbb{R})\). For all \(t\in \mathbb{R}\), \(f*\varphi (t)= \varphi*f(t)\).
We’ll remark here that these last two remarks are shadows of the theory of Fourier analysis. The convolution theorem (loosely) says that Fourier transform of a convolution of two functions is the product of the Fourier transforms of said functions. Then, the commutativity of the convolution is a consequence of the commutativity of multiplication of real numbers.
This does have ramifications in the preceeding proposition, however. One would be very tempted to say that \[\operatorname{supp} \left( \varphi* \psi \right) = \left\lbrace x+y : x\in \operatorname{supp}\varphi,\ y\in \operatorname{supp}\psi \right\rbrace;\] however, there are scenarios in which the inclusion is strict. In fact, it is very possible for the convolution of two nonzero functions to be zero (though we are not prepared to furnish explicit counterexamples).
Let’s wrap things up with a nice application of convolutions that emphasises the smoothing property.
Proposition 7.
Let \(f\in C \left( \mathbb{R}^2 \right)\) satisfy the mean value property: for any \(r > 0\) and \(x_0\in \mathbb{R}^2\), one has \[f\left( x_0 \right) = \frac{1}{2\pi r} \oint _{\partial B\left( x_0, r \right)} f(x) ds.\] Then, \(f\) is smooth.
Here, \(\partial B\left( x_0, r \right) = \left\lbrace x\in \mathbb{R}^2 : \left\lVert x-x_0 \right\rVert=r \right\rbrace\) is the boundary of the ball of radius \(r\) centred at \(x_0\).
Proof (Sketch)
Remark 8.
The above mean value property (together with continuity) is equivalent to being harmonic, which is a very powerful property. Harmonic functions play a central role in physics, differential equations, and complex analysis; in some sense, they are functions that are neither concave nor convex at any point in their domain.