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

出中七 Mock 卷,又是因為時間不足,沒有細心地把問題淺化。

談有關 convexity (凸性) 那道短題目。

考慮某區間 I,當

f'' \ge 0 on I – – – – – – (*)

我們熟知 the graph of y = f(x) is concave upwards。但對於 convex function,更基本的描述,可表如下

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" – – – – – – (**)

圖像意義是,在曲線 f 上任取兩點 A(x, f(x)), B(y, f(y)),則 chord AB 一定『高於』curve AB,見下:

注意到,(**) 是 concave upwards 的『更一般』條件,因為不像 (*),它所考慮的函數不需要兩次可導(second derivative exists)。

但若我額外加添 f'' 存在這一條件,相信我們可以證明

(*) <=> (**)

證 (*) => (**) 比較易,於是我想同學證 (**) => (*),這就是我出題的原意。

為簡化一些,我設 p = \frac{1}{2},得出以下提問。設 (a , b) 為非空區間。

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.

我之前做了一個所謂的 solution,見下:

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

上述的所謂證明是錯的,我又可以放它在『錯在哪裡』系列了。

其實我的所謂證明,問題(起碼)在於

第一,"taking limit y tends to x" 不一定可以,欲證 f'(x) increasing,我要在 (a , b) 任取兩個數 x , y,使 f'(y) \ge f'(x) for any y \ge x,而這兩個 x , y 不一定很接近。

第二,為何 \lim_{y \rightarrow x}\frac{f(y) - f(t)}{y - t}\lim_{t \rightarrow y}\frac{f(y) - f(t)}{y - t} 是一樣?似乎說明得不夠仔細,有點亂。

及後,我匆匆想了另一個證明。前半段和之前一樣。

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} – – – – – – (#)

之後,大概的想法是:由 t 出發,一路走到 y,我們可以得到很多 chords,由 (**),我們可知它們的 slopes 不斷上升,取極限,得 slope of chord = f'(y);如我們由 t 出發,一路走到 x,同樣地,我們可以得到很多 chords,由 (**),我們可知它們的 slopes 不斷下降,取極限,得 slope of chord = f'(x),於是有 f'(x) \le f'(y) 是也。

讓我把上述概念具體化,即

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.

看下圖,上述聲稱可謂顯而易見。

現在嘗試證明。

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.

如此,我們有

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

取極限 n \rightarrow \infty,即

\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)

循類似手法,我們可以構作 sequence of numbers {v_n},不斷接近 x,見下圖

得出 slope of AV_n 下降,從而
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)

再由 (#),得出
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。

因時間嚴重不足(要印卷了),我不打算改解答,反而改動題目,問較容易的方向 (*) => (**)。即同學現在所看到的 Q.6,即

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)

注意,其實我並未真正證明 (*) => (**),因為題目中的 x, y 是 fixed!! 我希望最終結論是 “for any x, y \in (a , b)“。同學,幫我 refine 吧。另外,part (b) 本來想問:有沒有滿足 (a) 的單射 (injective) 函數存在?最初,我還以為只有 linear function,及後我才知有無限可能性,都是要同學具體給例子這個問法更易,於是命之。

其實這個有關凸性的結果是很常見,只是我沒有深入看書,才想錯解答,發覺自己的學力是如此不濟,這裡大放厥詞,望高手勿笑。其他的題目,留待 part 2 再談。

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 | 回覆


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