3
$\begingroup$

I tried for 2 signals $s_1(t) = u(t)$ and $s_2(t) = (1+i) \cdot u(t)$. $i$ is the complex unit ($i = (-1) ^{0.5}$). I got the left hand side of the inequality as $(1-i)$ and the right hand side as $(1+i)^{0.5}$. The inequality is L.H.S $\le$ R.H.S.

How is inequality established for complex signals?

Improve this question
$\endgroup$
5
$\begingroup$

Schwarz Inequality for continuous-time Complex valued functions is given as follows: $$\left|\int^{\infty}_{-\infty}f(t)^* \cdot g(t) dt \right|^2 \le \int^{\infty}_{-\infty}\left|f(t)\right|^2dt \cdot \int^{\infty}_{-\infty}\left|g(t)\right|^2dt$$ As you can see that, on left hand side magnitude is taken after integration making it real number and the quantity on right hand side is also real number because magnitude-squared is taken before integrating. Thus, the comparison is valid and can be done for all Complex-valued continuous time functions which are Square integrable. Meaning the Inequality exists for all functions $f(t)$ satisfying the following: $$\int^{\infty}_{-\infty}\left|f(t)\right|^2dt = c < \infty$$

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks, I will edit. :) $\endgroup$ – DSP Rookie Apr 29 '20 at 9:19
2
$\begingroup$

$c$ need not be real number. For the LHS you would get $$ |\langle s_1,s_2\rangle| = |\langle s_1,cs_1\rangle |=|c|\,|\langle s_1,s_1\rangle |= |c|\|s_1\|^2 $$ For RHS, $$ \|s_1\| \times \|c s_1\| =\|s_1\| \times |c|\|s_1\| = |c|\|s_1\|^2 $$ So $|c|=|1+i|=\sqrt{2}$

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks for the suggest @RodrigodeAzevedo. Corrected $\endgroup$ – jithin Apr 29 '20 at 6:24
  • $\begingroup$ I am not familiar with this derivation. I am learning from Simon Haykin's book. The derivation there uses integration. Could you explain in more detail what the operations you used are? $\endgroup$ – Shashank V M Apr 29 '20 at 6:48
  • 1
    $\begingroup$ @ShashankVM the integration is just to compute the inner product $\langle s_1,s_2 \rangle$. You can replace that with $\int s_1^*(t)s_2(t)dt$. Similarly $||s_1||^2$ will be $\int |s_1|^2dt$ $\endgroup$ – jithin Apr 29 '20 at 7:57
1
$\begingroup$

The answer about $c$ being complex has be given already. I would add information on "How is inequality established for complex signals?" Indeed, the inegality is very general, and works in spaces equipped with an inner product (aka "inner-product spaces"). It is a very important inequality, in so many forms, that a whole book is dedicated to it: The Cauchy-Schwarz Master Class. An Introduction to the Art of Mathematical Inequalities, Steele, J. M.

In lectures, people often state CS (which should be called Cauchy-Bunyakovsky-Schwarz) for real signals, because complexes frighten students, and some proofs only work in the real case. But believe me (or not), complex spaces are natural to that respect. The real case is only a consequence of the more generic complex case (because reals are complexes without imaginary part). Let us see a proof, and reasons. Here $\langle u, v \rangle$ denotes a (possibly complex) scalar, and the squared-norm is $\|u\|^2 = \langle u, u \rangle$. Since the norm is always positive, this is true for any vector, for instance vector $u - \lambda \cdot v$, for any $\lambda$ (real or complex). So we have:

\begin{align} 0 & \leq \| u - \lambda \cdot v \|^2 \\ & \leq \langle u, u \rangle - \langle \lambda \cdot v, u \rangle - \langle u,\lambda \cdot v \rangle + \langle \lambda \cdot v, \lambda \cdot v \rangle \\ & \leq \langle u, u \rangle - \lambda \langle v, u \rangle - \overline{\lambda} \langle u, v \rangle + \lambda \overline{\lambda} \langle v, v \rangle \\ 0 & \leq \|u\|^2 - \lambda \overline{\langle u, v \rangle} - \overline{\lambda} \langle u, v \rangle + \lambda \overline{\lambda} \|v\|^2 \end{align}

Now, on way to prove the CBS inequality is to check the above of specific $\lambda$ values. To get the intuition, suppose for a second that $\|v\|\neq 0$. Then, let us choose the specific $\lambda$: $\lambda = \langle u, v \rangle / \|v\|^2$. Why? Because it will have, in modulus, big values when $u$ and $v$ are (more of less) aligned (i.e., what you get in your case: $u=c\cdot v$). Hence:

\begin{align} 0 & \le \|u\|^2 - \frac{|\langle u, v \rangle|^2}{\|v\|^2} - \frac{|\langle u, v \rangle|^2}{\|v\|^2} + \frac{|\langle u, v \rangle|^2}{\|v\|^2} \\ 0& \le \|u\|^2 - \frac{|\langle u, v \rangle|^2}{\|v\|^2}. \end{align}

A little rearrangement yields your inequality, since:

$$0 \cdot \|v\|^2 \le \|u\|^2 \|v\|^2 - |\langle u, v \rangle|^2\,,$$ thus: $$ |\langle u, v \rangle|^2 \le \|u\|^2 \|v\|^2 \,.$$

@Dilip Sarwate may ask: now, does this remain true when $\|v\| = 0$? The answer is yes. I did not do the maths, but I think one can build a sequence of $v_n$ that converge (in norm) to $v$, and prove the result at the limit. Not very elegant, but practical.

Improve this answer
$\endgroup$
  • $\begingroup$ For continuous-time square-integrable functions, it is not the case that $\lVert v\rVert =0$ implies that $v = 0$, that is, some nonzero functions have $0$ norm. Such signals are said to be equivalent to the zero signal. So the proof can only show that $\lVert u - \lambda \cdot v\rVert =0$, that is, $u$ and $\lambda \cdot v$ are equivalent in the sense that their difference is a signal equivalent to the zero signal. This is not quite the same as claiming that $u = \lambda \cdot v$ which implies equality at al time instants. $\endgroup$ – Dilip Sarwate Apr 29 '20 at 17:03
  • $\begingroup$ Are the $=$ signs below the $\le$ bothering your reading? $\endgroup$ – Laurent Duval Apr 29 '20 at 17:12
  • $\begingroup$ Would you prefer that I redefine $=$ as "almost-everywhere"? $\endgroup$ – Laurent Duval Apr 29 '20 at 17:16
  • 1
    $\begingroup$ I have no quarrel with $$\text{If }u=\lambda v, ~\text{then equality holds in CBS Inequality}$$ but don't agree with $$\text{If equality holds in CBS Inequality, then }u=\lambda v.$$ For the latter statement to hold, we also need the assumption that $$\lVert v\rVert > 0$$ because the latter claim is not true if the equality holds because both sides are $0$. Note that your proof implicitly uses the fact that $\lVert v\rVert > 0$ without which the choice of $\lambda = \frac{\langle u,v\rangle}{\lVert v\rVert^2}$ is invalid. Put in "almost everywhere" if you like. $\endgroup$ – Dilip Sarwate Apr 30 '20 at 14:49
  • $\begingroup$ Got if (I think). I have been light on $\lv\| \neq 0$ assumption, I can correct than. But I don't think I did to claim that equality leads to "vectors are colinear". Let me rewrite a bit $\endgroup$ – Laurent Duval Apr 30 '20 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.