陶哲轩实分析(上)9.7及习题-Analysis I 9.7

介值定理。很短但很有用。

Exercise 9.7.1

Since f f f is a continuous function on [ a , b ] [a,b] [a,b], by the maximum principle, there’s x 1 , x 2 ∈ [ a , b ] x_1,x_2∈[a,b] x1,x2[a,b] such that
M = f ( x 1 ) , m = f ( x 2 ) M=f(x_1 ),\quad m=f(x_2 ) M=f(x1),m=f(x2)
If x 1 = x 2 x_1=x_2 x1=x2, then let c = x 1 c=x_1 c=x1 and the proof is over, assume x 1 < x 2 x_1x1<x2, then by Exercise 9.4.6, we have f [ x 1 , x 2 ] f_{[x_1,x_2]} f[x1,x2] a continuous function on [ x 1 , x 2 ] [x_1,x_2 ] [x1,x2], by Theorem 9.7.1, ∃ c ∈ [ x 1 , x 2 ] ⊆ [ a , b ] ∃c∈[x_1,x_2 ]⊆[a,b] c[x1,x2][a,b], s.t. f ( c ) = y f(c)=y f(c)=y.

Exercise 9.7.2

We let F ( x ) = f ( x ) − x F(x)=f(x)-x F(x)=f(x)x, by Proposition 9.4.9 and Exercise 9.4.6, F ( x ) F(x) F(x) is continuous on [ 0 , 1 ] [0,1] [0,1], since f f f has range [ 0 , 1 ] [0,1] [0,1], we know that f ( 0 ) ≥ 0 f(0)≥0 f(0)0 and f ( 1 ) ≤ 1 f(1)≤1 f(1)1, thus
F ( 0 ) = f ( 0 ) ≥ 0 , F ( 1 ) = f ( 1 ) − 1 ≤ 0 F(0)=f(0)≥0,\quad F(1)=f(1)-1≤0 F(0)=f(0)0,F(1)=f(1)10
Since F ( 0 ) ≥ 0 ≥ F ( 1 ) F(0)≥0≥F(1) F(0)0F(1), by the Intermediate value theorem, there exists c ∈ [ 0 , 1 ] c∈[0,1] c[0,1] such that F ( c ) = 0 F(c)=0 F(c)=0, or f ( c ) = c f(c)=c f(c)=c, this is the fixed point we search for.

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