Re: [分析] 兩題
※ 引述《cckk3333 (皓月)》之銘言：
: 2. Let f: R -> R be a bounded function. Prove that f is continuous if and
: only if the graph = { (x,y) | y = f(x) } of f is a closed subset of R^2
Proof.
Suppose f is a continuous function defined on |R, we define F＝(I,f):|R→|R^2,
then F is continuous on |R. So, the graph is clearly a closed subset of |R^2.
Conversely, if the graph G_f is closed, we will show that f is continuous at a,
where a is given. Consider the graph which is contained in the compact region
D＝{(x,y): x in [a-1,a+1], y in [-M,M]}, where M ＝ sup {|f(x)|: x in |R}. So,
the restricted graph, called G'_f, is also compact. Let P:G'_f→[a-1,a+1] by
P(x,f(x))＝x. It is obvious that P is continuous and one-to-one on the compact
set G'_f. Hence, its inverse function P^-1 ＝ (I,f):[a-1,a+1] → G'_f is also
continuous (＊). It implies that f is continuous on [a-1,a+1]. In particular,
f is continuous at a. Since a is arbitrary, we have proved that f is continuous
on |R.
NOTE.
(1) The assumption that f is bounded is needed, e.g.,
f(x) ＝ 1/x if x ＞ 0,
＝ 0 if x ＝ 0,
＝ 1/x if x ＜ 0.
(2) For (＊), Walter Rudin, Principles of Mathematical Analysis, 3rd ed.
McGraw-Hill, 1976. Theorem 4.17, Chap. 4, p. 90.
(3) Reference: Walter Rudin, Principles of Mathematical Analysis, 3rd ed.
McGraw-Hill, 1976. Exercise 6, Chap. 4, p. 99.
(4) Related Exercise: Let f:(0,1)→|R. If the graph G_f is path-connected,
then f is continuous on (0,1).
(5) If you are interested with this, you can google "closed graph theorem"
in Functional Analysis.
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※ 編輯: math1209 來自: 140.113.25.169 (01/19 12:01)
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01/19 13:47, , 1F
01/19 13:47, 1F
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