# Quod Erat Demonstrandum

## 2008/02/26

### Something to say about F.7 Pure Mathematics (II) Mock Exam (part 1)

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:08 下午

$f'' \ge 0$ on I – – – – – – (*)

$f$ is concave upwards on I" means

$pf(x) + (1 - p)f(y) \ge f(px + (1 - p)y)$ for all $p \in [0 , 1]$ and $x , y \in I$" – – – – – – (**)

(*) <=> (**)

Suppose $f:(a , b) \rightarrow \mathbb{R}$ such that $f''(x)$ exists on $(a , b)$ and $\frac{f(x) + f(y)}{2} \ge f(\frac{x + y}{2})$ for all $x, y \in (a , b)$.
(a) Show that $f'(x)$ is increasing on $(a , b)$ and hence $f''(x) \ge 0$ on $(a , b)$.
(b) Suppose further that $f$ is injective on $(a , b)$, suggest an example of $f$.

For any $x , y \in (a , b)$, with $x < y$,
consider $t = \frac{x + y}{2}$,
by the condition given in question,
$\frac{f(x) + f(y)}{2} \ge f(t)$, then
$\frac{f(y) - f(t)}{y - t} \ge \frac{f(t) - f(x)}{t - x}$ – – – – – – (#)
Then taking limit y tends to x,
$\lim_{y \rightarrow x}\frac{f(y) - f(t)}{y - t} \ge \lim_{y \rightarrow x}\frac{f(t) - f(x)}{t - x}$
$\lim_{t \rightarrow y}\frac{f(y) - f(t)}{y - t} \ge \lim_{t \rightarrow x}\frac{f(t) - f(x)}{t - x}$
$f'(y) \ge f'(x)$
$\therefore f'(x)$ is increasing on $(a , b)$
hence $f''(x) \ge 0$

For any $x , y \in (a , b)$, with $x < y$,
consider $t = \frac{x + y}{2}$,
by the condition given in question,
$\frac{f(x) + f(y)}{2} \ge f(t)$, then
$\frac{f(y) - f(t)}{y - t} \ge \frac{f(t) - f(x)}{t - x}$ – – – – – – (#)

let {$u_n$} be a sequence of numbers such that $u_0 = t$, $u_{n} = \frac{u_{n-1} + y}{2}$ ($n = 1,2,...$).
let $U_n = (u_n , f(u_n))$ ($n = 1,2,...$).

Claim: slope of $BU_n$ increases as $n$ increases.

By the condition in the question, we have
$f(u_{n-1}) + f(y) \ge 2f(\frac{u_{n-1} + y}{2})$
$f(u_{n-1}) + f(y) \ge 2f(u_{n})$
$f(u_{n-1}) + 2f(y) \ge 2f(u_{n}) + f(y)$
$2[f(y) - f(u_{n})] \ge f(y) - f(u_{n-1})$
$\frac{f(y) - f(u_{n})}{y - u_{n}} \ge \frac{f(y) - f(u_{n-1})}{2(y - u_{n})}$
$\frac{f(y) - f(u_{n})}{y - u_{n}} \ge \frac{f(y) - f(u_{n-1})}{y - u_{n-1}}$
i.e. slope of $BU_n \ge$ slope of $BU_{n-1}$ as claimed.

slope of $BT \le$ slope of $BU_n$ for all positive \integers $n$.

$\lim_{n \rightarrow \infty}\frac{f(y) - f(t)}{y - t} \le \lim_{n \rightarrow \infty}\frac{f(y) - f(u_{n})}{y - u_{n}}$
$\frac{f(y) - f(t)}{y - t} \le \lim_{n \rightarrow \infty}\frac{f(y) - f(u_{n})}{y - u_{n}}$
$\frac{f(y) - f(t)}{y - t} \le \lim_{u \rightarrow y^-}\frac{f(u) - f(y)}{u - y} = f'(y)$$\because f'(y)$ exists）

slope of $AT \ge$ slope of $AV_n$ for any $n \in \mathbb{N}$
$f'(x)$ 的存在，順推
$\frac{f(t) - f(x)}{t - x} \ge f'(x)$

$f'(y) \ge \frac{f(y) - f(t)}{y - t} \ge \frac{f(t) - f(x)}{t - x} \ge f'(x)$

$f'(y) \ge f'(x)$ for $y \ge x$

$f$ 的遞增性 (increasing) 得證。

let $f:(a , b) \rightarrow \mathbb{R}$ such that $f''(x) \ge 0$ for $x \in (a , b)$.
For any $t \in (a , b)$, fixing $x$ and $y \in (a , b)$ with $x \le t \le y$. Take $t = px + (1 - p)y$ where $p \in [0 , 1]$.
Define $g(t) = pf(t) + (1 - p)f(y) - f(pt + (1 - p)y)$.
(a) Show that $g'(t) \le 0$ for any $t \in (a , b)$ and hence $pf(x) + (1 - p)f(y) \ge f(px + (1 - p)y)$.
(b) Suppose further that $f$ is injective, suggest an example of $f$.
(7 marks)

P.S.

Actually, my method above was really STUDPID! Here gives a better one.

Prove “slope of left chord $\le$ slope of right chord" $\rightarrow f'(x)$ is increasing".

Proof

$f'(x)$
$= \lim_{t\rightarrow x^+}\frac{f(t) - f(x)}{t - x}$
$\le \lim_{t\rightarrow x^+}\frac{f(t) - f(y)}{t - y}$
$= \frac{f(x) - f(y)}{x - y}$
$= \lim_{t\rightarrow y^-}\frac{f(x) - f(t)}{x - t}$
$\le \lim_{t\rightarrow y^-}\frac{f(y) - f(t)}{y - t}$
$= f'(y)$

No stupid sequences required.

Prove “$f'' \ge 0 \rightarrow$ “slope of left chord $\le$ slope of right chord".

Proof

For any $x,y \in (a , b)$, with $x < t < y$, by MVT, $\exists u , v$, with $x < u < t < v < y$, such that

$\frac{f(t) - f(x)}{t - x} = f'(u) \le f'(v) \le \frac{f(y) - f(t)}{y - t}$

Q.E.D.

Thank you my old self!

## 1 則迴響 »

1. 這一條….. 空了……

迴響 由 Wong Hon — 2008/02/26 @ 10:48 下午 | 回應