Question

Assume a = a(t) ∈ L1(0,1)∩C(0,∞), ∞ ≥ a(0+) > a(∞) ≥ 0 and a(t) ≥ 0, a′(t) ≤ 0, a′′(t) ≥ 0, a′′′(t) ≤ 0, on (0,∞). Let h(t) = ta(t) on (0,∞). Then one of the following is true. There exists a number ε > 0 such that h is increasing on (0, ε). Or, there is no such interval. Which is correct? Prove it.

Answer 1

Answer on Question # 41714 – Math – Real Analysis

Assume $a = a(t) \in L1(0,1) \cap C(0,\infty)$, $\infty \geq a(0+)$ > $a(\infty) \geq 0$ and $a(t) \geq 0$, $a'(t) \leq 0$, $a''(t) \geq 0$, $a'''(t) \leq 0$, on $(0,\infty)$. Let $h(t) = ta(t)$ on $(0,\infty)$. Then one of the following is true. There exists a number $\varepsilon$ > such that $h$ is increasing on $(0,\varepsilon)$. Or, there is no such interval. Which is correct? Prove it.

Solution.

$h(t) \in L_1(0,1) \cap C(0;\infty)$, as a composition of $a(t)$ and continuous function $t$. $h(t)$ has the same order of derivatives as $a(t)$.

Consider the derivative of $h(t)$:

$$h'(t) = \left(t \cdot a(t)\right)' = a(t) + t a'(t).$$

We obtain that $t a'(t) \leq 0$, as $t > 0$ and $a'(t) \leq 0$ on $(0;\infty)$.

So, $h(t)$ is increasing on interval $(0,\varepsilon)$ if $a(t) + t a'(t) > 0$ on it. As function $a(t)$ is decreasing and $t$ is increasing on $(0;\infty)$ there exist $t \in (0,\varepsilon) \subset (0;\infty)$: $a(t) > t a'(t)$.

For example, consider $a(t) = e^{-t}$ on $(0,\infty)$. $a(t) \in L_1(0,1) \cap C^\infty(0,\infty)$. And

$$a'(t) = -e^{-t} \leq 0,$$ $$a''(t) = e^{-t} \geq 0, a'''(t) = -e^{-t} \leq 0 \text{ on } (0,\infty).$$ $$h(t) = t e^{-t} \text{ on } (0;\infty).$$ $$h'(t) = e^{-t} - t e^{-t} > 0 \text{ for } t \in (0,\varepsilon), \quad \text{here } \varepsilon = 1.$$ $$h(t) = t e^{-t}:$$

Answer. There exists a number $\varepsilon \geq 0$ such that $h$ is increasing on $(0,\varepsilon)$.

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