Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

Question

Consider the function $f(x) = 1/x$ on the interval $I = [a, b]$, where $0 < a \leq b$. By Jensen's inequality, we have for any $\{x_j \}_{j=1}^n \subset I$,

$$ f(\overline{x}) \leq \frac{1}{n} \sum_{i=1}^n f(x_i). $$

Above, $\overline{x} = (1/n)\sum_{j=1}^n x_j$. Is there a converse inequality of the form $$ \frac{1}{n} \sum_{i=1}^n f(x_i) \leq C f\Big(\frac{1}{n}\sum_{i=1}^n x_i\Big), $$ where $C$ may depend on $a, b$?

One thing I thought of is a second-order Taylor expansion. By Taylor expansion, there are $\xi_i$ in the interval between $x_i, \overline{x}$ such that $$ \frac{1}{n} \sum_i f(x_i) = f(\overline{x}) + \frac{1}{2n} \sum_{i=1}^n f''(\xi_i) (x_i - \overline{x})^2. $$ Additionally, we have that $f''(\xi) = \tfrac{2}{\xi^3}$, so that the remainder is $$ \frac{1}{n} \sum_{i=1}^n \frac{(x_i - \overline{x})^2}{\xi_i^3} \leq \frac{1}{n} \sum_{i=1}^n \max \Big\{ \frac{(x_i - \overline{x})^2}{x_i^3}, \frac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}. $$ I struggled to bound this remainder in terms of $C(a, b)~f(\overline{x})$, though.

Answer 1

$C = \frac{(a+b)^2}{4ab}$ is an upper bound for the ratio, and that bound is best possible.

Proof: If $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ are real numbers in the interval $[m, M]$, $0 < m < M$, then $$ \sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2 \le \left(\dfrac{M+m}{2\sqrt{Mm}}\sum_{k=1}^n a_kb_k\right)^2 \, . $$ This is a discrete version of a Pólya-Szegő inequality, see for example Reverse Cauchy Schwarz for integrals.

We can apply this with $a_k = \sqrt {x_k}$, $b_k = 1/\sqrt{x_k}$, $[m,M] = [a, b]$. It follows that $$ \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k} \le n^2 \frac{(a+b)^2}{4ab} $$ and therefore $$ \frac 1n \sum_{k=1}^n f(x_k) \le \frac{(a+b)^2}{4ab} f\left( \frac 1 n \sum_{k=1}^n x_k\right) \, . $$

The bound is best possible: If $n$ is even then equality holds if half of the $x_k$ are equal to $a$ and the other half are equal to $b$. If $n$ is odd then asymptotic equality holds if $(n-1)/2$ of the $x_k$ are equal to $a$ and $(n+1)/2$ of the $x_k$ are equal to $b$.

Answer 2

Yes, there is a bound available.

Let $t_i := \max \Big\{ \tfrac{(x_i - \overline{x})^2}{x_i^3}, \tfrac{(x_i - \overline{x})^2}{\overline{x}^3} \Big\}$. We have that the remainder is the average $\frac{1}{n} \sum_{j=1}^n t_j$.

Consider the two cases separately. First suppose $x_i > \overline{x}$. Then $$ t_i = \frac{(x_i - \overline{x})^2}{\overline{x}^3} \leq \frac{1}{\overline{x}} \frac{x_i^2}{\overline{x}^2} \leq (b/a) \frac{1}{\overline{x}} \frac{x_i}{\overline{x}} $$

On the other hand, if $\overline{x} > x_i$, $$ t_i = \frac{(x_i - \overline{x})^2}{x_i^3} \leq \frac{1}{\overline{x}} \frac{\overline{x}^3}{x_i^3} \leq (b/a)^3 \frac{1}{\overline{x}}. $$ Therefore, $$t_i \leq (b/a)^3 \frac{1}{\overline{x}} + (b/a) \frac{1}{\overline{x}} \frac{x_i}{\overline{x}}$$.

If we average over $i=1, \dots, n$, we get $$ \frac{1}{n}\sum_{i=1}^n f(x_i) \leq f(\overline{x}) + (b/a)^3 f(\overline{x}) + (b/a) f(\overline{x}), $$ so that we may take $C(a, b) = 1 + (b/a)^3 + (b/a)$.

Answer 3

One can take $C(a, b) = b/a = 1 + (b-a)/a$.

To see this, note that $$ \frac{\frac{1}{n}\sum_i x_i^{-1}}{((1/n)\sum_{i=1}^n x_i)^{-1}} = \frac{1}{n^2} \Big(\sum_{i} x_i\Big) \Big(\sum_i x_i^{-1}\Big) \leq \frac{b}{a}. $$

Consider $x_1 = a, x_2 = b$. Then $$ \frac{\frac{1}{2} \frac{a + b}{ab}}{\frac{2}{a + b}} = \frac{(a+b)^2}{4ab} \geq \frac{1}{4} \frac{b}{a}. $$ Thus, this bound is optimal up to a constant $1/4$.

Answer 4

We want to determine an upper bound for the function $$ h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k} = \sum_{k, l=1}^n \frac{x_k}{x_l} $$ on the hypercube $[a, b]^n$. $h$ is convex in each variable, so that the maximum of $h$ is attained at a corner point of the hypercube. Therefore it suffices to compute $$ \begin{align} h(\underbrace{a, \ldots, a}_{m},\underbrace{b, \ldots, b}_{n-m}) &= \frac{1}{n^2}\bigl(ma - (n-m)b\bigr)\left( \frac m a + \frac{n-m} {b}\right) \\ &= \left( \frac mn \right)^2 + \frac mn\left( 1-\frac mn \right)\left( \frac a b + \frac b a\right) + \left( 1-\frac mn \right)^2 \\ &= \dots \\ &= \frac{(a+b)^2}{4ab} - \frac{(a-b)^2}{4ab}\left(2 \frac mn - 1\right)^2 \, . \end{align} $$

The sharp upper bounds are therefore $$ h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab} $$ if $n$ is even, and $$ h(x_1, \ldots, x_n) \le \frac{(a+b)^2}{4ab} - \frac{1}{n^2}\frac{(a-b)^2}{4ab} $$ if $n$ is odd.

The best upper bound which is independent of $n$ is $C = \frac{(a+b)^2}{4ab}$.