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Is it true that:

Let $\rho(\tau)\in (0,1]$. Suppose $\displaystyle \lim_{t\rightarrow\infty} \int_0^t \rho(\tau)y(\tau)^2\textrm{d}\tau$ is finite but $\displaystyle \lim_{t\rightarrow \infty} \int_0^ty(\tau)^2\textrm{d}\tau= \infty$. Then

  1. $\rho(t)\rightarrow 0$ as $t\rightarrow \infty$, provided $\rho(t)$ is a smooth function.

Secondly, what are the conditions on $\rho(t)$ if it is not a smooth function?

For (1), I start by taking the lower bound of the first integral: $\displaystyle \left(\lim_{t\rightarrow\infty} (\inf_t{\rho(t)})\right) \infty\leq \lim_{t\rightarrow\infty}{\left( (\inf_t{\rho(t)})\int_0^t y(\tau)^2\textrm{d}\tau\right)}\leq \lim_{t\rightarrow\infty} \int_0^t \rho(\tau)y(\tau)^2\textrm{d}\tau<\infty$.

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    Do you mean $fg\in L^1,$ and $t\to \infty$ in the title? (The conclusion looks very doubtful by the way) – 2017-02-27
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    Hi Zhw. To the former, I did mean $fg \in L_2$, i.e. since $f\in(0,1]$, $\int_0^\infty f^2 g^2\textrm{d}\tau\leq \int_0^\infty f g^2\textrm{d}\tau=\ell< \infty$. To the latter, yes, thank very much you for catching that. I've updated it. – 2017-02-27

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This is false. Integral hypotheses on $\rho$ don't control the pointwise behavior of $\rho$, even if it's smooth. (Integrals involving the derivative of $\rho$ could have such control, though.)

For example, let $y\equiv 1$ and choose $\rho(\tau) = \frac12\exp(-\tau) + \frac12 (\sin^2 \tau)^{\exp(\tau)}$. The nonsensical formula aside, the point is that $\rho$ has extremely narrow peaks, which prevent it from having limit zero, yet are too narrow to add up to much in terms of the integral.

A more sensible example would be $y(\tau)=1/\sqrt{\tau}$ and $\rho$ being the sum of characteristic functions of intervals $[2^n, 2^n+1]$; here the computations can be carried out explicitly and then $\rho$ can be smoothened, made positive, and seasoned to taste.