Equality of $L_2$ norm on $[0, \infty)$ Statement:
For a continuous function $y(t)$ defined on $[0,\infty)$, 
if $\int_0^{\infty}\frac{y^2(\tau)}{1+y^2(\tau)}d\tau<\infty$, $\int_0^{\infty}y^2(\tau)<\infty$.
The statement is true if $y(t)\in L_{\infty}$. What about if $y(t)$ is bounded for any finite time $t$? Is the above statement still true?

You can find continuous functions $y_n$, with support in $[0,1/n^2]$ such that $\int_0^{\infty}y_n^2(\tau) = n$.  And of course, $\int_0^{\infty}\frac{y_n^2(\tau)}{1+y_n^2(\tau)}d\tau \le 1/n^2$.
Now let $y(x) = y_n(x-n)$ if $n \le x < n+1$ ($n \in \mathbb N$), and you have your counterexample.
