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

 A: 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 
$$\int_{T}|x(t)| < \infty$$
A: A periodic continuous function is bounded, so both integrals are necessarily finite.
A: 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.
A: 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 $p \leq q$?
