Is true the following assertion?

Question

$\forall f\in C^\infty(\Bbb R):\quad \int^\infty_0 \big(16t^2|f''(t)|^2+t^2|f(t)|^2-8t^2|f'(t)|^2\big)dt>0$

any counter-example will be appreciate.

Answer 1

No, it is not true. Take $f(t) = \exp(t)$ as a smooth counterexample.

We have:

• $f(t) = \exp(t) \in C^{\infty}$
• $\exp(t) > 0$ for all $t \in \mathbb{R}$
• $f(t) = f'(t) = f''(t) = \exp(t)$

So your integral

$$ I_1 = \int^\infty_0 t^2|f''(t)|^2+t^2|f(t)|^2-8t^2|f'(t)|^2 dt $$

can for $f(t) = \exp(t)$ be simplified to

$$ \begin{align} I_1 &= \int^\infty_0 t^2 \exp(t)^2 + t^2 \exp(t)^2 - 8 t^2 \exp(t)^2 dt \\ &= -6 \int^\infty_0 t^2 \exp(t)^2 dt \rightarrow -\infty \end{align} $$

so $I_1$ approaches negative infinity.

Edit: Considering your updated question, you have

$$ I_2 = \int^\infty_0 16 t^2|f''(t)|^2+t^2|f(t)|^2-8t^2|f'(t)|^2 dt $$

and exclude the zero function explicitly. As stated in a comment, $I_2 = 0$ for $f(t) = \exp(t/2)$. However, I don't think that even the relaxed inequality $I_2 \geq 0$ is true. Consider

$$ f(t) = (t + 1)^{-0.4 t} $$

With this $f$, the exact value of the integral is very hard to calculate, but using a numeric solver, the value is

$$ I_2 \approx -0.1668 $$

Alternatively, for a function that is defined on all $\mathbb{R}$, take

$$ f(t) = (t^2/10 + 1)^{-0.4 t} $$

for which you get $I_2 \approx -0.1588$. Of course, by scaling $f$ with a constant, you can make the integral even "more negative".

I doubt that the statement is true, even with your new assumptions.