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陶哲轩实分析10.3及习题-Analysis I 10.3

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Exercise 10.3.1. Prove Proposition 10.3.1.
Proposition 10.3.1 Let X be a subset of \mathbf R, let x_0\in X be a limit point of X, and let f:X\to \mathbf R be a function. If f is monotone increasing and f is differentiable at x_0, then f'(x)\geq 0. If f is monotone decreasing and f is differentiable at x_0, then f'(x)\leq 0.

Solution: Since f is differentiable at x_0, we have

f'(x_0 )=\lim\limits_{x\to x_0;x\in X-{x_0 }} \dfrac{f(x)-f(x_0)}{x-x_0}

If f is monotone increasing, then we have

f(x)\leq f(x_0 ),x<x_0,\quad\quad f(x)\geq f(x_0 ),x>x_0

Thus in both x<x_0 and x>x_0 we shall have

\dfrac{f(x)-f(x_0)}{x-x_0}\geq 0 \implies f'(x_0 )=\lim\limits_{x\to x_0;x\in X-{x_0 } }\dfrac{f(x)-f(x_0)}{x-x_0}\geq 0

The case when f is monotone decreasing can be similarly proved.

\blacksquare

Exercise 10.3.2. Give an example of a function f:(-1,1)\to\mathbf R which is continuous and monotone increasing, but which is not differentiable at 0. Explain why this does not contradict Proposition 10.3.1.

Solution: Define

f(x)=\begin{cases}x,&x\in (-1,0]\\2x,&x\in (0,1) \end{cases}

This does not contradict Proposition 10.3.1 since Proposition 10.3.1 requires f to be differentiable at the point (0 in this case).

\blacksquare

Exercise 10.3.3. Give an example of a function f:\mathbf R\to\mathbf R which is strictly monotone increasing and differentiable, but whose derivative at 0 is zero. Explain why this does not contradict Proposition 10.3.3.

Solution: Define

f(x)=x^3,\quad x\in (-1,1)

Then f is differentiable at 0, f'(0)=0, but f is monotone increasing.
This doesn’t contradict Proposition 10.3.1 since f'>0 is a necessary but not sufficient condition for f to be strictly monotone increasing.

\blacksquare

Exercise 10.3.4. Prove Proposition 10.3.3.
Proposition 10.3.3. Let a<b, and let f:[a,b]\to\mathbf R be a differentiable function. If f'(x)>0 for all x\in [a,b], then f is strictly monotone increasing. If f'(x)<0 for all x\in [a,b], then f is strictly monotone decreasing.If f'(x)=0 for all x\in [a,b], then f is a constant function.

Solution: For any x\neq y\in [a,b], without loss of generality we can suppose x<y, then f|_{[x,y]} is continuous and differentiable on [x,y], thus by mean value theorem, we can find a c\in (x,y)\subset (a,b) such that

f'(c)=\dfrac{f(y)-f(x)}{y-x}=\dfrac{f(x)-f(y)}{x-y}

If f'(x)>0,\forall x\in [a,b], then f'(c)>0 and f(x)<f(y), so f is strictly monotone increasing.
If f'(x)<0,\forall x\in [a,b], then f'(c)<0 and f(x)>f(y), so f is strictly monotone decreasing.
If f'(x)=0,\forall x\in [a,b], then f'(c)=0 and f(x)=f(y), so f is a constant function.

\blacksquare

Exercise 10.3.5. Give an example of a subset X\subset \mathbf R and a function f:X\to\mathbf R which is differentiable on X, is such that f'(x)>0 for all x\in X, but f is not strictly monotone increasing.

Solution: Define

f(x)=\begin{cases}x+1,&x\in (-1,0)\\x-1,&x\in (0,1)\end{cases}

Then if X=(-1,0)\cup (0,1), then f is differentiable on X and f'(x)=1>0,\forall x\in X, but we have f(1/2)=-1/2<f(-1/2)=1/2, thus f in not strictly monotone increasing.
The key condition which is different from Proposition 10.3.3 is that X is allowed to be a disconnected set.

\blacksquare