Boundedness of integral I am currently working on a research project and have kind of stuck in proving the boundedness of a certain integral of a continuously differentiable signal $f(t)$. I have managed to prove the following properties:


*

*$f(t)$ is uniformly bounded.

*$\frac{1}{t}\int_0^t{sf^2(s)}ds$ is uniformly bounded.
From 2. it is easy to prove that
$$\inf_{\tau\in[0,t]}|f(\tau)|\leq \sqrt{\frac{2\sigma}{t}}\qquad\qquad \qquad\qquad(1)$$
for some $\sigma>0$.
From the simulations I run it seems that the whole function has a behavior of the form
$$|f(t)|\leq \sqrt{\frac{\eta}{t}}\qquad\qquad \qquad\qquad\qquad\quad(2)$$
for some $\eta>0$ but I could not prove that. 
So my question is this: 
Do these properties 1., 2. suffice to prove that $\int_0^t{f^4(\tau)d\tau}$ is uniformly bounded or do I need to impose some extra assumption?
Note that from (2) the boundeness of $\int_0^t{f^4(\tau)d\tau}$ follows directly. 
Note also that I cannot use boundedness of $f'(t)$ since, in my current setting, this property follows as a consequence of the boundedness of the integral $\int_0^t{f^4(\tau)d\tau}$.
 A: You need to impose some further assumptions. I will ignore the differentiability and give a merely continuous example for simplicity, but it should be clear that mollification to obtain a smooth function doesn't alter the overall behaviour.
Let $g$ be a piecewise linear function, such that it has a symmetric triangular spike of height $C > 0$ and width $\frac{2}{n}$ on the interval $[n - 1/n, n + 1/n]$ for $n \geqslant 2$, and let it be $0$ elsewhere. Let $f(\tau) = \sqrt{g(\tau)}$. Then $f$ is continuous, and uniformly bounded - $0 \leqslant f(\tau) \leqslant \sqrt{C}$ for all $\tau$ - and we have
$$\int_{n - 1/n}^{n + 1/n} sf(s)^2\,ds = n\int_{n-1/n}^{n+1/n} g(s)\,ds = n\cdot \frac{C}{2}\cdot \frac{2}{n} = C$$
for $n \geqslant 2$. It follows that
$$\lim_{t\to \infty} \frac{1}{t} \int_0^t s f(s)^2\,ds = C.$$
But, for $n \geqslant 2$ we have
$$\int_{n-1/n}^{n+1/n} f(\tau)^4\,d\tau = 2\int_0^{1/n} (Cns)^2\,ds = \frac{2C^2}{3n},$$
whence
$$\int_0^t f(\tau)^4\,d\tau \sim \frac{2C^2}{3}\log t.$$
