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I wrote down a proof for the following proposition, and it is quite different from the standard proof which views the integral as a function of upper limit. Can anyone please verify it for me?

Proposition: Assume $f(x)$ is continuous and increasing on the inteval $[a,b]$, prove that $$\int_a^b (\frac{b-x}{b-a})^n f(x) dx \leq \frac{1}{n+1}\int_a^b f(x)dx.$$

My proof: The original inequality is equivalent to the following one $$\int_a^b [(n+1)(b-x)^n-(b-a)^n] f(x)dx \leq 0.$$ Let $g(x) = (n+1)(b-x)^n-(b-a)^n$, and notice that $g(x)$ is exactly the derivative of the function $G(x) = (b-x)[(b-a)^n - (b-x)^n]$. So it is sufficient to show $$\int_a^b G'(x)f(x)dx \leq 0.$$

We prove the inequality above using the definition of definite integral. Given a partition $$a=x_0\leq x_1\leq x_2\leq\dots\leq x_n \leq x_{n+1} = b$$ of $[a,b]$ into $n$ subintevals $[x_j,x_{j+1}]$ of same length $\frac{b-a}{n}$, $j=1,2,\dots,n$.

It is easy to see $G(x) \geq 0$ for all $x \in [a,b]$, and $f(x_k)-f(x_{k-1}) \geq 0$ since $f(x)$ is increasing. It follows that $$\sum_{k=0}^n G(x_{k+1})[f(x_{k+1})-f(x_k)] \geq 0.$$ By Abel's identity we have $$\sum_{k=0}^nf(x_k)[G(x_{k+1})-G(x_k)] \\ = f(x_{k+1})G(x_{k+1})-f(x_0)G(x_{k+1})-\sum_{k=0}^n G(x_{k+1})[f(x_{k+1})-f(x_k)] \\ = -\sum_{k=0}^n G(x_{k+1})[f(x_{k+1})-f(x_k)] \leq 0 ,$$ where the last equality follows from $G(x_{k+1}) = G(b) = 0$ and $G(x_{0}) = G(a) = 0$.

By the definition of differentiation, $G(x_{k+1})-G(x_k) = G'(x_k)(x_{k+1} - x_k) + o(x_{k+1} - x_k) = g(x_k)(x_{k+1} - x_k) + o(\frac{b-a}{n})$, thus the inequaltity $$\sum_{k=0}^nf(x_k)[G(x_{k+1})-G(x_k)] \leq 0$$ becomes $$\sum_{k=0}^nf(x_k)g(x_k)(x_{k+1} - x_k) + \sum_{k=0}^nf(x_k)o(\frac{b-a}{n}) \leq 0.$$ Now let $n \to \infty$ and the first term of the left hand side becomes $\int_a^b g(x)f(x)dx$ and the second term becomes $0$, which completes the proof.

zyy
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Your proof looks correct to me. What you essentially have done is integration by parts with the Riemann-Stieltjes integral: $$ \int_a^b G'(x)f(x)dx = \int_a^b f(x)dG(x) = -\int_a^b G(x) df(x) \ge 0 \, , $$ see for example Riemann-Stieltjes integral, integration by parts (Rudin).

A simple alternative proof would be to apply the “Integral Chebyshev inequality” to the function $f$ and $$ h(x) = \left(\frac{b-x}{b-a}\right)^n \, . $$ $f$ and $h$ are of opposite monotonicity, so that $$ \int_a^b h(x) f(x) \, dx \le \frac{1}{b-a} \int_a^b h(x)\, dx \int_a^b f(x) \, dx = \frac{1}{n+1} \int_a^b f(x) \, dx \, . $$

You can also prove it directly by noting that $$ \int_a^b \int_a^b (f(x)-f(y))(h(x)-h(y)) \, dx dy \le 0 $$ and expanding the left-hand side.

Martin R
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  • Thanks. I did not know the concept of Riemann-Stieltjes Integral, but I did notice that the inequality follows easily if we assume $f$ to be differentiable and integrate $\int_a^b g(x)f(x)dx$ by part, so I have to use alternatively this 'discrete version of integration by part'. – zyy Jul 01 '22 at 17:08
  • @zyy: Yes, that is what I assumed. I just wanted to mention that it can also be done with R-S integrals if $f$ is just continuous (and the proof of IBP with R-S integrals resembles what you have done, Abel summation etc.) – Martin R Jul 01 '22 at 17:17
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    Thank you for letting me know the principle behind this proof, and also teaching me a new concept which seems quite intriguing to me. – zyy Jul 01 '22 at 17:26