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Is convolution a linear operator?

  1. Nov 26, 2009 #1
    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?
  2. jcsd
  3. Nov 26, 2009 #2


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    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]".)
  4. Nov 26, 2009 #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: Nov 26, 2009
  5. Nov 27, 2009 #4


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    Yes, and that is a special case (n=2) of (n)-multilinearity.
  6. Dec 7, 2009 #5
    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: Dec 7, 2009
  7. Dec 7, 2009 #6


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    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: Dec 7, 2009
  8. Dec 7, 2009 #7
    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 seperate 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".
  9. Dec 8, 2009 #8


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    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.
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