Why is it not true $\int_{T}|x(t)|^2 < \infty \implies $ $\int_{T}|x(t)| < \infty $ for a periodic function $x(t)$?

Question

Why is it not true $\int_{T}|x(t)|^2 < \infty \implies $ $\int_{T}|x(t)| < \infty $ for a periodic continuous function $x(t)$ with period $T$ ?

This is coming straight from my textbook, see the part below "but not vice versa"

The premise of the question is false -- this statement is true. — Caleb Stanford, Mar 15 '20 at 01:55

score 2 · Answer 1 · answered Mar 15 '20 at 01:14

2

By Cauchy-Schwarz you have $$\left( \int_{T} |x(t)|d t \right)^2 \leq \left( \int_{T} |x(t)|^2d t \right)\left( \int_{T} 1^2 d t \right)$$

Since $T$ has finite measure, it follows that $$\int_{T}|x(t)| < \infty$$

answered Mar 15 '20 at 01:14

N. S.

132,525

But I"m still not seeing why it is not true. Can you give me a counter example for x(t)? – user1068636 Mar 15 '20 at 01:51
@user1068636 They are saying it is true. – Caleb Stanford Mar 15 '20 at 01:54
@6005 - are you sure? This is coming straight from my textbook. See the picture I added above. – user1068636 Mar 15 '20 at 01:58
@user1068636 It's probably one of those non standard examples people use outside mathematics, which $T$ could be infinite. This is very common in signal analysis and engineering. An examples was provided. – ShoutOutAndCalculate Mar 15 '20 at 02:00
Yes, this is a digital signal processing textbook. So what is the counter example? – user1068636 Mar 15 '20 at 02:01
@user1068636 See the example provided above, when $T$ be infinite, the proof no longer valid and is basically a calculus II question. You could replace $1/t$ with $ sin(1/t)$ as well. – ShoutOutAndCalculate Mar 15 '20 at 02:03
@user1068636 Periodic means that the period is finite. This would not be true over an infinite region, and there are standard counterexamples. – N. S. Mar 15 '20 at 02:07
@N.S. I knew, that's why I said "non standard mathematics", they do not strictly assume $T$ necessarily to be of finite measure in order to complete some proofs. – ShoutOutAndCalculate Mar 15 '20 at 02:10
@user1068636 Yes I am 110% sure, but you have now changed the question -- the book, unlike your original question, does not assume $x(t)$ is continuous. In that case the book appears to be right. – Caleb Stanford Mar 15 '20 at 04:30

score 2 · Answer 2 · answered Mar 15 '20 at 01:16

2

A periodic continuous function is bounded, so both integrals are necessarily finite.

answered Mar 15 '20 at 01:16

Karl

11,446

score 2 · Accepted Answer · answered Mar 15 '20 at 04:50

Two things are happening here:

When you copied the book's question here, you added the assumption that $x(t)$ is continuous which was not present in the book. That changes the result.
The book has seemingly made a typo. The actual implication is the other way around: (4.17) implies (4.16), but not vice versa.

This is all very confusing, so let me first answer the question you literally asked, and then let me explain what the book was trying to say (but it made a typo).

Why is it not true $\int_{T}|x(t)|^2 < \infty \implies $ $\int_{T}|x(t)| < \infty $ for a periodic continuous function $x(t)$ with period $T$ ?

Emphasis added. This statement is not true, because of the continuous assumption. In fact, as Karl correctly points out, the assumption that $x(t)$ is continuous implies that both integrals are necessarily finite.

Proof: Since $x(t)$ is continuous on all of $\mathbb{R}$, it must be continuous on the closed interval $[a, a+T]$ for some $a$, which is a closed and bounded (compact) interval, and a continuous function on a compact set is bounded. Then if $|x(t)|$ is bounded by $M$ this implies the upper bounds $\int_{T}|x(t)|^2 \le T M^2$ and $\int_{T}|x(t)| < T M$, so both integrals are finite.

In your book, however, it appears that $x(t)$ may be discontinuous. However, it is assumed to have finitely many discontinuities, and importantly, I guess these could be asymptotes, so the function may be unbounded. So here is what your book wanted to say:

Claim: $\int_{T}|x(t)|^2 < \infty$ implies $\int_{T}|x(t)| < \infty$, but not vice versa.

N.S. has already given the proof of the implication, which uses Cauchy Schwartz. For the counterexample, note the following examples:

$\int_0^1 \frac{1}{x^{0.9}}$ converges, but $\int_0^1 \frac{1}{x^{1.8}}$ does not.
In general, $\int_0^1 \frac{1}{x^p}$ converges whenever $0 < p < 1$, but diverges when $p \ge 1$. When you square it, that doubles $p$, so we may go from converging to diverging.

score 0 · Answer 4 · answered Mar 15 '20 at 01:48

Not sure, but technically make the period to infinity, and a very classical example was $x(t)=1/t$ non zero only for $t\in [1,\infty]$.

But this is just a technicality(when assuming $T$ could be infinite), there should be other better examples as well.

Check Related: How do you show that $l_p \subset l_q$ for $p \leq q$?

Why is it not true $\int_{T}|x(t)|^2 < \infty \implies $ $\int_{T}|x(t)| < \infty $ for a periodic function $x(t)$?

4 Answers4