陶哲轩实分析（上）10.2及习题-Analysis I 10.2

局部最值，Roll定理和中值定理（Lagrange）。

Exercise 10.2.1

First suppose $f$ attains a local maximum at $x_0$, then $∃δ>0$, s.t.
$f(x)≤f(x_0 ),\quad ∀x∈(a,b)∩(x_0-δ,x_0+δ)$
Since $f$ is differentiable at $x_0$, the limit
$\lim_{x→x_0;x∈(a,b))}\frac{f(x)-f(x_0)}{x-x_0}$
exists, thus both left and right limit exists and is equal to $f' (x_0 )$.
Notice that
$\frac{f(x)-f(x_0)}{x-x_0 }≥0,\quad x∈(a,b)∩(x_0-δ,x_0 )$
and
$\frac{f(x)-f(x_0)}{x-x_0 }≤0,\quad x∈(a,b)∩(x_0,x_0+δ)$
We shall have
$f' (x_0 )=\lim_{x→x_0;x∈(a,b)∩(x_0-δ,x_0 ) }⁡\frac{f(x)-f(x_0)}{x-x_0 }≥0$
and
$f' (x_0 )=\lim_{x→x_0;x∈(a,b)∩(x_0,x_0+δ) }\frac{f(x)-f(x_0)}{x-x_0 }≤0$
Thus $f' (x_0 )=0$.
The case when $f$ attains a local minimum at $x_0$ can be similarly proved.

Exercise 10.2.2

Define
$f(x)=-|x|,\quad x∈(-1,1)$
Then $f$ is continuous and attains a global maximum at $0$, but is not differentiable at $0$.
This doesn’t contradict Proposition 10.2.6, since Proposition 10.2.6 requires $f$ to be differentiable at local maximum or minimum.

Exercise 10.2.3

Define
$f(x)=x^3,\quad x∈(-1,1)$
Then $f$ is differentiable at 0, $f' (0)=0$, but $0$ is neither a local minimum nor local maximum.
This doesn’t contradict Proposition 10.2.6, since derivative equals zero is a necessary but not sufficient condition for a a local minimum nor local maximum.

Exercise 10.2.4

If $∀x∈(a,b),g(x)=g(a)=g(b)$, then $g$ is constant on $[a,b]$ and $g' (x)=0,x∈(a,b)$ by Theorem 10.1.13.
Now suppose if $∃x∈(a,b),g(x)≠g(a)$, then $g(x)≠g(b)$. Since $g$ is continuous on $[a,b]$ we can use the maximum principle, say $f$ attains it’s maximum at $x_{max}∈[a,b]$, attains it’s minimum at $x_{min}∈[a,b]$. We either have $g(x)>g(a)$ or $g(x)<g(a)$ by the trichotomy of real numbers.
If $g(x)>g(a)$, then $f$ can’t attain its maximum at a or b, so $x_{max}∈(a,b)$, since $x_{max}$ is also a local maximum, we have $f' (x_{max} )=0$ by Proposition 10.2.6.
If $g(x)<g(a)$, then $f$ can’t attain its minimum at $a$ or $b$, so $x_{min}∈(a,b)$, since $x_{min}$ is also a local minimum, we have $f' (x_{min} )=0$ by Proposition 10.2.6.
In conclusion the statement is true in all cases.

Exercise 10.2.5

We let
$F(y)=f(y)-\frac{f(b)-f(a)}{b-a} (y-a)$
Then by Proposition 9.4.9 and Theorem 10.1.13, $F(y)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $F(a)=f(a),F(b)=f(a)$, we can use Roll’s theorem to say $∃x∈(a,b),F' (x)=0$, or
$F' (x)=f' (x)-\frac{f(b)-f(a)}{b-a} =0 ⇒ f' (x)=\frac{f(b)-f(a)}{b-a}$

Exercise 10.2.6

For any $x,y∈[a,b]$, if $x=y$ the statement is obviously true. Without loss of generality we can suppose $x<y$, then $f|_{[x,y]}$ is continuous on $[x,y]$ and differentiable on $(x,y)$, thus by mean value theorem, we can find a $c∈(x,y)⊂(a,b)$ such that
$f' (c)=\frac{f(y)-f(x)}{y-x}=\frac{f(x)-f(y)}{x-y}$
Since $|f'(x)|≤M,∀x∈(a,b)$, we have
$|f'(c)|=\left|\frac{f(x)-f(y)}{x-y}\right|≤M⇒|f(x)-f(y)|≤M|x-y|$

Exercise 10.2.7

Since $f'$ is bounded, $∃M>0$, s.t. $|f'(x)|≤M,∀x∈\mathbf R$. Then given $∀ϵ>0$, let $δ=ϵ/M$, then as long as $|x-y|<δ$, we can get from Exercise 10.2.6 that
$|f(x)-f(y)|≤M|x-y|<Mδ=ϵ$
which means $f$ is uniformly continuous.