In other words, how can I show that if we feed an energy signal to a system whose impulse response is an energy signal, the output will also be an energy signal?

I've been trying using Rayleigh's energy theorem and Cauchy Schwarz inequality but didn't work out.

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    $\begingroup$ // if we feed an energy signal to a system whose impulse response is an energy signal, the output will also be an energy signal?// I don't think it's true. Dimensionally, if the output of the LTI system is the same dimension of the input, then the impulse response of a continuous-time LTI system must have dimension of time${}^{-1}$ . $\endgroup$ Oct 14, 2021 at 7:20
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    $\begingroup$ This is 1:1 what is explained in en.wikipedia.org/wiki/BIBO_stability#Proof_of_sufficiency $\endgroup$ Oct 14, 2021 at 8:09
  • $\begingroup$ But @MarcusMüller, i think that this $\| h \|_1$ in $$ \int_{-\infty}^\infty \left|h(t)\right|\,\mathord{\operatorname{d}}t = \| h \|_1 < \infty $$ must be dimensionless. $\endgroup$ Oct 14, 2021 at 8:48
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    $\begingroup$ Dimensionful numbers can be meaningfully compared to either $0$ or $\infty$. But they can't be compared to $1$ or $e$ or $\pi$ or a non-zero real number. But $0$ meters has meaning. $\endgroup$ Oct 14, 2021 at 8:57
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    $\begingroup$ Maybe helpful: math.stackexchange.com/questions/118230/… $\endgroup$
    – S.H.W
    Oct 14, 2021 at 21:44

2 Answers 2


Potentially, no. I will attempt to find that $L_2$ is not closed under convolution. In other words, a convolution of two $L_2$ functions (or energy functions) is not necessarily in $L_2$. Convolution is a complicated operation, so working in the Fourier domain with products of functions is tempting. However, if Fourier provides an isometry between the primal time and the dual, it often requires proofs with tempered distributions. I am not sure I am still able to do that correctly.

Therefore, I will search for functions $f\in L_1 \cap L_2$ where Fourier behaves nicely for DSP people. Take $$f_a: x\mapsto \frac{e^{-x^2}}{|x|^a}\,.$$

If $ a <1$, $f_a$ is integrable (in $L_1$). If $2a<1$, it is square integrable (in $L_2$). If $4a > 1$, its square $f_a^2$ is not square integrable anymore (non integrable singularity at $0$). Therefore, for $\frac{1}{4} < a<\frac{1}{2}$, we have

  • $f_a \in L_1$,
  • $f^2_a \in L_1$,
  • $f_a \in L_2$,
  • $f^2_a \notin L_2$.

If we note $\hat{f_a}$ for the Fourier transform of $f_a $ (well defined), then $\hat{f_a}$ is square integrable, but $\hat{f_a}*\hat{f_a}$ is not.

The rest is for memory. What can be proved is that the convolution of two $L_2$ functions $x$ and $h$ is bounded. A sketch of the proof indeed uses the Cauchy-Schwarz inequality:

\begin{align}\left|\int_\mathbb{R} x(t'-t)h(t)\mathrm{d}t\right|& \le \int_\mathbb{R} \left| x(t'-t)\right|\left|h(t)\right|\mathrm{d}t\\&\le \left(\int_\mathbb{R} \left| x(t'-t)\right|^2\mathrm{d}t\right)^{1/2}\left(\int_\mathbb{R}\left|h(t)\right|^2\mathrm{d}t\right)^{1/2}\end{align} which is finite ($x*h\in L_\infty$) since $x$ and $h$ are $L_2$ or "energy signals". This shape of proof extends to functions in $L_p$ and $L_q$ respectively as long as $p$ and $q$ are conjugated:

$$ \frac{1}{p}+ \frac{1}{q}=1\,.$$

Side notes:

  • $L_1$ on the contrary is closed with convolution: if $x$ and $h$ are in $L_1$, $x*h$ is in $L_1$ as well.
  • if the LTI system $h$ is only absolutely integrable or BIBO ($h\in L_1$), and $x\in L_2$, the convolution still exists almost everywhere, and remains in $L_2$.
  • if one does not consider the real line anymore, like periodic functions, things might be different.
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    $\begingroup$ thank you very much, Laurent $\endgroup$ Oct 15, 2021 at 13:35
  • $\begingroup$ This question made me reconsider what I thouht was obvious. Appreciated $\endgroup$ Oct 26, 2021 at 5:56

Assuming by energy signal means $\int x^2(t)dt < \infty$ here is one way to do it.

Let's say we have

$$y(t) = x(t)*h(t)$$

Convolution in time is multiplication in frequency, so we have

$$Y(\omega) = H(\omega) \cdot X(\omega) $$

Due to Parseval's Theorem we have $$\int x^2(t)dt = \int |X(\omega)|^2 d \omega$$


$$\int y^2(t)dt = \int |X(\omega)|^2 \cdot |H(\omega)|^2 d \omega \leq \int |X(\omega)|^2 d \omega \cdot \int |H(\omega)|^2 d \omega $$

Since $x(t)$ and $h(t)$ have finite energy, so has $y(t)$.

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    $\begingroup$ I don't think that inequality that you used in the last line is quite right $\endgroup$ Oct 14, 2021 at 15:19

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