Help to understand finding a bound to this integral

Question

• How do we use integration by parts to get the formula $$\int_0^{\infty} x^\gamma f(x)dx = \frac{1}{1+\gamma}\int_0^{\infty} x^{\gamma+1}|f^{'}(x)|dx?$$ I think I understand this part now. He claimed this because he also assumed $x^{\gamma+1}f(x)\to 0$ as $x\to\infty$.
• How does $x^{\gamma+1}f(x)\to 0$ as $x\to\infty$? $f$ is nonincreasing by supposition. What if $f$ is a constant function?
• How does $1-\left(\frac{\alpha-\beta}{\alpha+\beta+1}\right)^2 = \frac{(2\alpha+1)(2\beta+1)}{(\alpha+\beta+1)^2}$? My calculation gives $1-\left(\frac{\alpha-\beta}{\alpha+\beta+1}\right)^2<\frac{(2\alpha+1)(2\beta+1)}{(\alpha+\beta+1)^2}$

---

Answer 1

The integration by parts is as follows: \begin{align} \int_0^\infty \underbrace{x^\gamma }_{dv}\; \underbrace{f(x)}_{u}\;dx &= \underbrace{f(x)}_{u}\underbrace{\tfrac{1}{1+\gamma}x^{\gamma+1}}_{v}\big]_0^\infty - \int_0^\infty \underbrace{\tfrac{1}{1+\gamma}x^{\gamma+1}}_{v}\underbrace{f'(x)\;dx}_{du}\\ &= \frac{1}{1+\gamma}\left(\lim_{x\to \infty} f(x)x^{\gamma+1} - \lim_{x\to 0^+} f(x) x^{\gamma+1} + \int_0^{\infty} x^{\gamma+1}\big(-f'(x)\big)\;dx\right)\\ &= \frac{1}{1+\gamma}\left(\underbrace{0}_{(1)} - \underbrace{f(0)\cdot 0}_{(2)} + \int_0^{\infty} x^{\gamma+1}\underbrace{|f'(x)|}_{(3)}\;dx\right)\\ &= \frac{1}{1+\gamma} \int_0^\infty x^{\gamma+1}|f'(x)|\;dx \end{align}

I'll leave (1) till the end. (2) follows because $f$ is continuous on $[0,\infty)$ so $f(0)$ is well defined and we can separate the limit. (3) follows as $f$ is non-increasing and differentiable, hence $f'(x)\leq 0$, so $-f'(x) = |f'(x)|$.

As to your next question: \begin{align}\frac{(2\alpha+1)(2\beta+1)}{(\alpha+\beta + 1)^2} &= 1-\left(1-\frac{(2\alpha+1)(2\beta+1)}{(\alpha+\beta + 1)^2}\right) = 1-\frac{(\alpha+\beta + 1)^2 - (2\alpha+1)(2\beta+1)}{(\alpha+\beta + 1)^2}\\ &= 1-\frac{\big(\alpha^2+2\alpha\beta+\beta^2+2\beta(1)+(1)^2 + 2(1)\alpha\big) - \big(4\alpha\beta+2\alpha + 2\beta + 1\big)}{(\alpha+\beta+1)^2}\\ &= 1 - \frac{a^2-2\alpha\beta + \beta^2}{(\alpha+\beta+1)^2} = 1 - \left(\frac{\alpha-\beta}{\alpha+\beta+1}\right)^2 \end{align}

Now (1) is addressed at the bottom part of the proof you posted, but I will try and spell out each and every painstaking detail:

Our solution is therefore complete except for one small point; we need to check that our three applications of the integration by part formula (7.5) were justified. For this it suffices to show that we have the limit (7.6) when $\gamma$ equals $2\alpha$, $2\beta$, or $\alpha+\beta$,

That is we want to confirm $$\lim_{x\to\infty}x^{\alpha+\beta+1}f(x)\;dx =\lim_{x\to\infty}x^{2\alpha+1}f(x)\;dx = \lim_{x\to\infty}x^{2\beta+1}f(x)\;dx = 0, $$

and it clearly suffices to check the limit for the largest of these,

This is because $$0\leq x^{2\alpha+1}f(x),\;x^{2\beta+1}f(x), \;x^{\alpha+\beta+1}f(x)\leq x^{\max\{2\alpha,2\beta,\alpha+\beta\}+1}f(x) $$ so the squeeze theorem shows that if the right hand side goes to $0$ as $x\to \infty$ then all of the limits are equal to $0$.

which we can take to be $2\alpha$.

Note that the inequality we're trying to prove is symmetric in $\alpha$ and $\beta$. In other words swapping $\alpha$ and $\beta$ leads to the same inequality, so if we show it for $\alpha \geq \beta$ then we have also shown it for $\beta \geq \alpha$.

So if $\alpha \geq \beta$ (and we continue this assumption hereafter) then

$$2\beta = \underbrace{\beta}_{\leq\;\alpha}+\beta \leq \alpha+\underbrace{\beta}_{\leq\;\alpha} \leq 2\alpha = \max\{2\alpha,\alpha+\beta,2\beta\}$$

Moreover, we can assume that in addition to the hypotheses of the challenge problem that we also have the condition $$\int_0^\infty x^{2\alpha}f(x)\;dx<\infty,$$ since otherwise our target inequality (7.4) is trivial

I feel that the author poorly presented the general idea here. Essentially the point is that if $$\int_0^\infty x^{2\alpha}f(x)\;dx=\infty$$ then the inequality is equivalent to Cauchy-Schwartz.

Cauchy-Schwartz gives $$I^2 \leq \underbrace{\int_0^\infty x^{2\alpha}f(x)\;dx}_{\equiv \; A} \underbrace{\int_0^\infty x^{2\beta}f(x)\;dx}_{\equiv \; B}$$ which holds even when the integrals are infinite. Note that although we don't know whether or not $B$ is infinite, we certainly know it is greater than $0$, as $A=\infty$ and $f(x)\geq 0$. (Rigorously there must exist some interval on which $f(x)>0$, so the integration multiplied by any non-negative function is non-zero over this interval).

Note that we also don't know whether or not $I^2 = \infty$, but this doesn't matter as $$I^2 \leq \underbrace{A}_{\infty} \cdot \underbrace{B}_{> 0} = \infty $$ so $$I^2 \leq \eta \infty = \eta AB = \infty$$ for any $\eta > 0$. In particular, this holds when $\eta = 1 - \left(\frac{\alpha-\beta}{\alpha+\beta+1}\right)^2$. (This is all rigorous in terms of Extended real numbers).

Hence Cauchy Schwartz implies our inequality "trivially". Otherwise we have $$\int_0^\infty x^{2\alpha} f(x) \;dx < \infty$$ and we need to use (7.6)

Unfortunately here the author completely skips over why this implies $\lim_{x\to\infty} x^{2\alpha+1} f(x) = 0$. This follows because if we had $\ell = \lim_{x\to\infty} x^{2\alpha+1} f(x) > 0$ (the limit must exists as $f$ is bounded below and non-increasing.) As $f$ is non-increasing it follows that $x^{2\alpha+1}f(x) \geq \ell$, hence \begin{align} \int_0^\infty x^{2\alpha} f(x) \;dx &= \int_0^1 x^{2\alpha} f(x) \;dx + \lim_{t\to\infty}\int_1^t \frac{1}{x}x^{2\alpha+1} f(x) \;dx\\ &\geq \int_0^1 x^{2\alpha} f(x) \;dx + \lim_{t\to\infty}\int_1^t \frac{1}{x} \ell \;dx\\ &= \int_0^1 x^{2\alpha} f(x) \;dx + \ell \lim_{t\to\infty}\log(t) = \infty \end{align} so $\int_0^\infty x^{2\alpha} f(x) \;dx=\infty$. Hence $\ell = 0$, i.e, (7.6) holds.