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Is the energy preserved in convolution?

I convolve two functions:

$$g(t) = f_1(t) \cdot f_2(t)$$

provided that the integrals of $f_1$ and $f_2$ remain unchanged, is the integral of $g(t)$ always the same, independent on the temporal shape of $f_1$ and $f_2$?

grg
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user3004286
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    Define "energy". The integral $\int g = (\int f_1 )(\int f_2)$ as a consequence of Fubini. But if you consider energy as something else, then possibly not. – Willie Wong Feb 18 '15 at 15:34

1 Answers1

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Yes, using Fubini's theorem, it is not hard to see that

\begin{eqnarray*} \int (f_1 \ast f_2)(x) \, dx &=& \int \int f_1(y) f_2(x-y) \, dy \, dx \\ &=& \int f_1(y) \int f_2(x-y) \, dx \, dy \\ &=& \int f_1(y) \int f_2 (z) \, dz \, dy \\ &=& \int f_1(y) \, dy \cdot \int f_2 (z) \, dz, \end{eqnarray*} for $f_1, f_2 \in L^1$.

But the word energy is usually used for the (squared) $L^2$-norm, i.e.

$$ \Vert f \Vert_2^2 = \int |f|^2 \, dx. $$

This is not preserved, since the Plancherel theorem yields

$$ \Vert f_1 \ast f_2 \Vert_2^2 = \Vert \mathcal{F}(f_1 \ast f_2)\Vert_2^2= \Vert \widehat{f_1} \cdot \widehat{f_2} \Vert_2^2, $$ and the right hand side vanishes if $\widehat{f_1} \cdot \widehat{f_2} \equiv 0$. This can happen even if $\Vert f_1 \Vert_2 \neq 0 \neq \Vert f_2 \Vert_2$, e.g. if $\widehat{f_1} = 1_{[0,1]}$ and $\widehat{f_2} = 1_{[42, 100]}$.

PhoemueX
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