Is convolution a linear operator?

In summary, the conversation discusses the concept of linearity in the context of convolution operators and functions of multiple complex vectors. It is established that a function can be considered linear in its components if it satisfies certain conditions, and that this definition can be extended to include conjugate linearity. The conversation also explores different definitions of convolution and how to properly format equations in a forum setting.
  • #1
mnb96
715
5
Hello,
If f is a morphism between two vector spaces, we say it is linear if we have:
1) [itex]f(x+y) = f(x) + f(y)[/itex]
2) [itex]f(ax) = af(x)[/itex]

Now, if f is the convolution operator [itex]\ast[/itex] , we have a binary operation, because convolution is only defined between two functions.

Can we still talk about linearity in this case or it does not make sense?
In case it makes sense, what does the definition of a binary linear operator looks like?
 
Physics news on Phys.org
  • #2
You can talk about linearity in each "component". That is, if f(x,y)= v, then f is linear in its first component if and only if f(au+ bv, y)= af(u,y)+ bf(v,y). It is linear in its second component if and only if f(x, au+ bv)= af(x,u)+ bf(x,v). It may be said to be "linear" if it is linear in both components.

(If our vector space is over the complex numbers, the condition for "linear in the second component may becomes
"[tex]f(x, au+ bv)= \overline{a}f(x,u)+ \overline{b}f(x,v)[/tex]".)
 
  • #3
Ok, that was very clear, thanks!
So I shall conclude that since convolution is distributive under addition, it must be then linear in both arguments.
One last question: were did you find these definitions? The books I have always deal with the basic definition, and never consider more general cases like this.

edit: was that actually the definition of bilinearity? (if it works for both the components)
 
Last edited:
  • #4
mnb96 said:
edit: was that actually the definition of bilinearity? (if it works for both the components)

Yes, and that is a special case (n=2) of (n)-multilinearity.
 
  • #5
HallsofIvy said:
(If our vector space is over the complex numbers, the condition for "linear in the second component may becomes
"[tex]f(x, au+ bv)= \overline{a}f(x,u)+ \overline{b}f(x,v)[/tex]".)

Do you mean that in general, for a function of n complex vectors, multilinear means "linear or conjugate linear (antilinear) in each of its arguments", or if not, how is multilinearity defined for a function of n complex vectors?

I see that the inner product of a complex vectors space is conjugate linear in its second argument, which follows from linearity of the first argument and the requirement that

[tex]\left \langle \mathbf{u},\mathbf{v} \right \rangle = \overline{\left \langle \mathbf{v},\mathbf{u} \right \rangle}[/tex]

But is it possible to have a function of at least two complex vectors that's just linear in its second argument?
 
Last edited:
  • #6
Maybe our definitions of convolutions are different, but the definition I have learned for the convolution of two functions is defined by an integral, i.e. $ f*g(t)=\int_{0}^{t}f(u)g(t-u)du.$ Letting $v=t-u,$ then $dv=-du$($t$ is constant with respect to $u$). We also need to change our limits of integration. For $u=0$ we have $v=t,$ and for $u=t$ we have $v=0.$ Thus $f*g(t)=-\int_{t}^{0}g(v)f(t-v)dv=\int_{0}^{t}g(v)f(t-v)dv.$ The final expression is $g*f(t).$ The fact that we used $v$ instead of $u$ is unimportant.


Note: this is my first post and I apologize for not being able to enter the equations such that they appear as they would in LaTeX.
 
Last edited:
  • #7
crd said:
Maybe our definitions of convolutions are different, but the definition I have learned for the convolution of two functions is defined by an integral, i.e.

[tex] f*g(t)=\int_{0}^{t}f(u)g(t-u)du.[/tex]

Letting [tex]v=t-u,[/tex] then [tex]dv=-du[/tex] ([tex]t[/tex] is constant with respect to [tex]u[/tex]). We also need to change our limits of integration. For [tex]u=0[/tex] we have [tex]v=t,[/tex] and for [tex]u=t[/tex] we have [tex]v=0.[/tex] Thus

[tex]f*g(t)=-\int_{t}^{0}g(v)f(t-v)dv=\int_{0}^{t}g(v)f(t-v)dv.[/tex]

The final expression is [tex]g*f(t).[/tex] The fact that we used [tex]v[/tex] instead of [tex]u[/tex] is unimportant.


Note: this is my first post and I apologize for not being able to enter the equations such that they appear as they would in LaTeX.

Hi crd, and welcome to Physics Forums. You just need to put "tex" in square brackets before the LaTeX and "/tex" in square brackets after it. Unfortunately the LaTeX isn't always lined up well with the text of the text. At least when I look at it in Firefox, the LaTeX symbols sometimes appear raised above the line of the text slightly. And bigger symbols, such as the integral signs can look very uneven with big gaps between the lines where they appear, so I usually put them on a separate line so that it's easier to read. There are also some Physics Forums tags you can use such as "i" in square brackets followed by any text you want in italics, then "/i". For bold "b" and "/b". For superscript, "sup" and "/sup". And for subscript, "sub" and "/sub".
 
  • #8
Rasalhague said:
Unfortunately the LaTeX isn't always lined up well with the text of the text. At least when I look at it in Firefox, the LaTeX symbols sometimes appear raised above the line of the text slightly. And bigger symbols, such as the integral signs can look very uneven with big gaps between the lines where they appear, so I usually put them on a separate line so that it's easier to read.
Use [noparse][itex]...[/itex][/noparse] instead of [noparse][tex]...[/tex][/noparse] for inline tex, i.e. LaTeX within a paragraph of text. It's rendered smaller and more accurately aligned.
 

FAQ: Is convolution a linear operator?

1. What is a convolution operator?

A convolution operator is a mathematical operation that combines two functions to produce a third function. It is commonly used in signal and image processing to manipulate and analyze data.

2. Is convolution a linear operator?

Yes, convolution is a linear operator. This means that it satisfies the properties of linearity, such as the superposition principle and homogeneity. In other words, the output of a convolution operation is directly proportional to the input.

3. How does convolution work?

Convolution works by multiplying two functions, one of which is flipped and shifted, and then integrating the product over a specific range. This process is repeated for different shifts, resulting in a new function that represents the combined effect of the two original functions.

4. What are the applications of convolution?

Convolution has various applications in fields such as signal and image processing, mathematics, and engineering. It is used for smoothing and filtering data, edge detection, and pattern recognition, among others.

5. Are there different types of convolution?

Yes, there are different types of convolution, such as discrete convolution, continuous convolution, and circular convolution. Each type has its own specific properties and applications.

Back
Top