$\int_0^\infty f(t)^2 x(t)^2 dt \le \int^1_0 f(t)^2 dt?$

Question

Let $f$ let a bounded decreasing positive function on $(0,\infty)$, and $x$ be a positive function such that $x(t)\le 1$ for every $t\in (0,\infty)$ and $\int_0^\infty (x(t))^{2}dt\le 1$. Show (or disprove) that $$ \int_0^\infty f(t)^2 x(t)^2 dt \le \int^1_0 f(t)^2 dt.$$

Answer 1

Write $g(t) = f(t)^2$ and $y(t) = x(t)^2$ and notice that

• $g$ is bounded decreasing non-negative on $(0, \infty)$.
• $y$ is non-negative, bounded above by $1$ and $\int_{0}^{\infty} y(t) \, dt \leq 1$.

So we have

\begin{align*} \int_{1}^{\infty} g(t)y(t) \, dt &\leq g(1) \int_{1}^{\infty} y(t) \, dt \\ &\leq g(1) \left(1 - \int_{0}^{1} y(t) \, dt \right) \\ &= \int_{0}^{1} g(1) (1 - y(t)) \, dt \\ &\leq \int_{0}^{1} g(t) (1 - y(t)) \, dt. \end{align*}

Rearranging, we obtain

$$ \int_{0}^{\infty} g(t)y(t) \, dt \leq \int_{0}^{1} g(t) \, dt $$

which is equivalent to the inequality in question.