Exercise 10, § 1

Functions of a Real Variable, III, 117

The sequence $(a_k)_{k \geq 1}$ being formed of numbers $> 0$ , we put $s_n = n A_n = a_1 + \cdots + a_n$ . For $p < 1$ and $\lambda_k > 0$ one has

\mu_1 s_1^{1/p} + \mu_2 s_2^{1/p} + \cdots + \mu_n s_n^{1/p} \leq \lambda_1 a_1^{1/p} + \lambda_2 a_2^{1/p} + \cdots + \lambda_n a_n^{1/p}

provided that $\mu_1 \geq \lambda_1$ and, for $k \geq 2$ , $\mu_k = \left( \lambda_k^q + \delta_k^q \right)^{1/\delta} - \delta_{k+1}$ , with $q = p/(1-p)$ , $\delta_k \geq 0$ for every $k$ , and $\delta_{n+1} = 0$ (method of exerc. 8). In particular, one has the inequalities

\sum_{k=1}^n \left( (k-p)^{1/q} - k^{1/q} \right) s_k^{1/q} + n^{1/q} s_n^{1/q} \leq (1-p)^{1/q} \sum_{k=1}^n a_k^{1/q}

which imply (with $A_n = s_n/n$ )

\sum_{k=1}^n A_k^{1/p} + \frac{n}{1-p} A_n^{1/p} \leq (1-p)^{-1/p} \sum_{k=1}^n a_k^{1/p}

(strengthened Hardy inequality);

\frac{1}{q} \sum_{k=1}^n (b_k - 1) A_k^{1/p} + n A_n^{1/p} \leq \sum_{k=1}^n a_k^{1/p} b_k^{1/p}

provided that $b_k \geq 1$ for all $k$ ;

\alpha(a_1^2 + a_2^2 + \cdots + a_n^2) \leq a_1^2 + (a_2 - a_1)^2 + \cdots + (a_n - a_{n-1})^2 + \beta a_n^2

\alpha = 2(1 - \cos \theta)

and

\beta = 1 - \frac{\sin(n+1)\theta}{\sin n\theta}

for some

\theta

such that

0 \leq \theta < \frac{\pi}{n}

Case when $\theta = \frac{\pi}{n+1}$ ; case when $\theta = \frac{\pi}{2n+1}$ .

Solution

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