Convolution Notation: The difference between (f*g)(x) and f(x)*g(x)











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What is the difference between (f*g)(x) and f(x)*g(x) [1] for convolutions? Are they the same? I ask this because I have been asked to prove the Reflection of Convolution property for my course in the Theory of Distributions,
i.e that f(-x)g(-x)=(fg)(-x). But if there is no difference between [1], then surely the proof follows from [1] and nothing actually needs to be proven.






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    What is the difference between (f*g)(x) and f(x)*g(x) [1] for convolutions? Are they the same? I ask this because I have been asked to prove the Reflection of Convolution property for my course in the Theory of Distributions,
    i.e that f(-x)g(-x)=(fg)(-x). But if there is no difference between [1], then surely the proof follows from [1] and nothing actually needs to be proven.






    notation convolution reflection







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    user617486 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
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      up vote
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      down vote

      favorite











      What is the difference between (f*g)(x) and f(x)*g(x) [1] for convolutions? Are they the same? I ask this because I have been asked to prove the Reflection of Convolution property for my course in the Theory of Distributions,
      i.e that f(-x)g(-x)=(fg)(-x). But if there is no difference between [1], then surely the proof follows from [1] and nothing actually needs to be proven.






      notation convolution reflection







      share|cite|improve this question







      New contributor




      user617486 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      What is the difference between (f*g)(x) and f(x)*g(x) [1] for convolutions? Are they the same? I ask this because I have been asked to prove the Reflection of Convolution property for my course in the Theory of Distributions,
      i.e that f(-x)g(-x)=(fg)(-x). But if there is no difference between [1], then surely the proof follows from [1] and nothing actually needs to be proven.






      notation convolution reflection




      notation convolution reflection






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          $f*g$ means that $*$ is defined on the space of functions. The notation $f(x)*g(x)$ makes it look as though $*$ is defined on the space of real numbers (if the functions are real that is), and hence isn't great. It's the same solecism as talking about 'the function $f(x)$'---it's $f$ that is the function, not $f(x)$.






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            $f*g$ means that $*$ is defined on the space of functions. The notation $f(x)*g(x)$ makes it look as though $*$ is defined on the space of real numbers (if the functions are real that is), and hence isn't great. It's the same solecism as talking about 'the function $f(x)$'---it's $f$ that is the function, not $f(x)$.






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              up vote
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              $f*g$ means that $*$ is defined on the space of functions. The notation $f(x)*g(x)$ makes it look as though $*$ is defined on the space of real numbers (if the functions are real that is), and hence isn't great. It's the same solecism as talking about 'the function $f(x)$'---it's $f$ that is the function, not $f(x)$.






              share|cite|improve this answer























                up vote
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                up vote
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                down vote









                $f*g$ means that $*$ is defined on the space of functions. The notation $f(x)*g(x)$ makes it look as though $*$ is defined on the space of real numbers (if the functions are real that is), and hence isn't great. It's the same solecism as talking about 'the function $f(x)$'---it's $f$ that is the function, not $f(x)$.






                share|cite|improve this answer












                $f*g$ means that $*$ is defined on the space of functions. The notation $f(x)*g(x)$ makes it look as though $*$ is defined on the space of real numbers (if the functions are real that is), and hence isn't great. It's the same solecism as talking about 'the function $f(x)$'---it's $f$ that is the function, not $f(x)$.







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                Richard Martin

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