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Vince
Vince
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I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correctEDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

Apologies, we can check this formula viaonly applies in the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$case that $f$ is compactly supported. Back to the drawing board...

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

Apologies, this only applies in the case that $f$ is compactly supported. Back to the drawing board...

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Vince
Vince
  • 505
  • 5
  • 14

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

Source Link
Vince
Vince
  • 505
  • 5
  • 14

I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.