### Chapter 5 Differentiation

#### Exercise 1

(By Matt Frito Lundy) For any $x$, we have:

\begin{align*}

\left| \frac{f(t) – f(x)}{t – x} \right| &\leq \left| \frac{(t – x)^2}{t – x} \right| \\

&=\left| t – x \right| \to 0

\end{align*}as $t \to x$. Because $f’(x) = 0$ for all $x$, $f(x)$ is a constant by Theorem 5.11(b).

#### Exercise 2

(By Matt Frito Lundy) If $f$ were not strictly increasing in $(a,b)$, there would exist $x,y$ with both $a < x < y < b$ and $f(x) \geq f(y)$. By *the* mean value theorem, there would exist a $z \in (x,y)$ such that $f’(z) = \frac{f(y) – f(x)}{y – x} \leq 0$, which contradicts $f’(x) >0$ in $(a,b)$, so $f$ must be strictly increasing.

Fix $\varepsilon > 0$ and $x \in (a,b)$. $f’(x) >0$ means there exists a $\eta > 0$ and a $\delta_1 >0$ such that $0 < \left| x – t \right| < \delta_1$ implies

$$\left| \frac{f(t) – f(x)}{t-x} \right| > \eta$$ So:

$$\frac{1}{\left| \frac{f(t) – f(x)}{t-x} \right|} < \frac{1}{\eta}$$From the definition of $f’(x)$, we also have for $\eta f’(x) \varepsilon > 0$, there exists a $\delta_2 > 0$ such that $0 < \left| x – t \right| < \delta_2$ implies

$$\left| f’(x) – \frac{f(t) – f(x)}{t-x} \right| < \eta f’(x) \varepsilon$$Let $y = f(x)$ and $\delta = \min \{\delta_1,\delta_2\}$. Then for any $u \in (f(x – \delta), f(x+\delta))$, let $g(u) = t$ so $t \in (x -\delta, x + \delta)$ and:

\begin{align*}

\left|\frac{g(u) – g(y)}{u – y} – \frac{1}{f’(x)} \right|

&= \left| \frac{f’(x) – \frac{f(t) – f(x)}{t-x}}{f’(x) \left[ \frac{f(t) – f(x)}{t-x} \right] } \right| \\

&< \frac{\eta f’(x) \varepsilon}{\eta f’(x)} \\

&= \varepsilon

\end{align*}Which shows that

$$g’(f(x)) = \frac{1}{f’(x)}.$$

#### Exercise 3

(By analambanomenos) The derivative of $f$ is $f’=1+\varepsilon g’$, which is positive if $\varepsilon<1/M$. In that case, by Exercise 2, $f$ is strictly increasing, hence one-to-one.

#### Exercise 4

(By Matt Frito Lundy) Let:

$$f(x) = \sum_{i = 0}^n \frac{C_i x^{i+1}}{i+1}.$$ Then we are looking for an $x \in (0,1)$ such that $f’(x) = 0$. But $f(x)$ is continuous in $[0,1]$, differentiable in $(0,1)$, $f(0) = 0$, and $f(1)= 0$. So by *the* mean value theorem, there exists an $x \in (0,1)$ such that $f’(x) = 0$, as desired.

#### Exercise 5

(By Matt Frito Lundy) Fix $\varepsilon > 0$. $f’(x) \to 0$ as $x \to + \infty$ means that there exists an $M \in \mathbf R$ such that $x \geq M$ implies $\left| f’(x) \right| < \varepsilon$. Now for any $a, b \in \mathbf R$ such that $M < a < b$, *the* mean value theorem implies that there exists an $x \in (a,b)$ such that

$$\frac{f(b) – f(a)}{b-a} = f’(x). $$ But $x \in (a,b)$ implies that $x > M$ so:

$$\left| \frac{f(b) – f(a)}{b-a} \right| = \left| f’(x) \right|

< \varepsilon.$$ Taking $b = a+1$ means if $a > M$, then $\left| g(a) \right| < \epsilon$, which shows that $g(x) \to 0$ as $x \to +\infty$.

#### Exercise 6

(By analambanomenos) By Theorem 5.10, for $x>0$ we have $f(x)=f(x)-f(0)=(x-0)f’(y)$ for some $y\in(0,x)$, so that $f(x)\le xf’(x)$ since $f’$ is monotonically increasing. Hence $$g’(x)=

\frac{xf’(x)-f(x)}{x^2}\ge 0\quad(x>0),$$ and so $g$ is monotonically increasing by Theorem 5.11(a).

#### Exercise 7

(By Matt Frito Lundy) We have:

$$ \lim_{t \to x} \frac{f(t)}{g(t)}

= \lim_{t \to x} \frac{f(t) – f(x)}{t – x} \frac{t – x}{g(t) – g(x)}

= \frac {f’(x)}{g’(x)}$$

#### Exercise 8

(By Matt Frito Lundy) Because $f’$ is continuous on the compact set $[a,b]$, $f’$ is uniformly continuous (Theorem 4.19). So there exists $\delta >0$ such that for any $x,y$ where $|x-y| < \delta$ we have $|f’(x) – f’(y)| < \varepsilon$. But by “the” mean value theorem, for any $x,t \in [a,b]$, there exists a $y \in (a,b)$ such that $|x – y| < |x -t|$ and

$$f’(y) = \frac{f(t) – f(x)}{t -x }.$$ So if $\left| x- t \right| < \delta$ we have:

$$\left| \frac{f(t) – f(x)}{t -x } – f’(x) \right| = \left| f’(y) – f’(x) \right| < \varepsilon$$ because $|x – y| < |x – t| < \delta$.

(By analambanomenos) Let $\mathbf f$ be a vector-valued function such that $\mathbf f’$ is continuous on $[a,b]$ and $\varepsilon>0$. Define the vector-valued function $\mathbf g$ on the rectangle $[a,b]\times[a,b]$ as follows:

$$\mathbf g(x,t)=

\begin{cases}

\displaystyle\frac{\mathbf f(t)-\mathbf f(x)}{t-x}-\mathbf f’(x) & x\neq t \\

\boldsymbol 0 & x=t

\end{cases}$$Then $\mathbf g$ is continuous on its compact domain, and so is uniformly continuous by Theorem 4.19. Hence there is $\delta>0$ such that $|\mathbf g(x,t)|=

\big|\mathbf g(x,t)-\mathbf g(x,x)\big|<\varepsilon$ if $\big|(x,t)-(x,x)\big|=|t-x|<\delta$.

#### Exercise 9

(By Matt Frito Lundy) By definition $$f’(0) = \lim_{t \to 0}\frac{f(t) – f(0)}{t}.$$Because $f$ is continuous on $\mathbf R$ and differentiable on $\mathbf R \setminus 0$, *the* mean value theorem says for any $t \in \mathbf R \setminus 0$, there exists an $x_t \in (0,t)$ (or $(t,0)$ if $t$ is negative) such that

$$\frac{f(t) – f(0)}{t} = f’(x_t).$$ So we have:

$$f’(0) = \lim_{t \to 0}\frac{f(t) – f(0)}{t} = \lim_{t \to 0} f’(x_t) = 3.$$ Because $x_t \to 0$ as $t \to 0$. So $f’$ exists at $0$, and is $3$.

#### Exercise 10

(By analambanomenos) Let $f(x)=u(x)+iv(x)$ where $u,v$ are real-valued and differentiable on $(0,1)$, and let $A=a+ib$. Following the hint, since $$\frac{u’(x)}{1}\rightarrow a\quad

\hbox{and}\quad\frac{v’(x)}{1}\rightarrow b\quad\hbox{as $x\rightarrow 0$,}$$ we have by Theorem 5.13 $$\frac{u(x)}{x}\rightarrow a\quad\hbox{and}\quad

\frac{v(x)}{x}\rightarrow b\quad \hbox{as $x\rightarrow 0$}$$ so that $$\frac{f(x)}{x}\rightarrow A\quad\hbox{as $x\rightarrow 0$}.$$ Similarly, by breaking $g$ into its real and imaginary parts, we get $$\frac{g(x)}{x}\rightarrow B\quad\hbox{and so}\quad\frac{x}{g(x)}\rightarrow\frac{1}{B}\quad\hbox{as $x\rightarrow 0$}.$$ Hence

$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0} \Biggl(\biggl(\frac{f(x)}{x}-A\biggr)\frac{x}{g(x)}+A\frac{x}{g(x)}\Biggr)=\frac{A}{B}.$$In Example 5.18 it was shown that the derivative of the denominator $g$ satisfies $\big|g’(x)\big|\ge (2/x)-1$. Since the limit of $g’(x)$ as $x\rightarrow 0$ does not exist, we can’t apply the above result to this case.

#### Exercise 11

(By analambanomenos) Let $g(h)=f(x+h)+f(x-h)-2f(x)$. Then $g$ is differentiable in a neighborhood of 0, and $g(0)=0$. Applying Theorem 5.13, we get

\begin{align*}

\lim_{h\rightarrow 0}\frac{g(h)}{h^2} &= \lim_{h\rightarrow 0}\frac{f’(x+h)-f’(x-h)}{2h} \\

&=\lim_{h\rightarrow 0}\frac{f’(x+h)-f(x)}{2h}-\lim_{k\rightarrow 0}\frac{f(x+k)-f(x)}{-2k}\quad\hbox{(where $k=-h$)} \\

&=\frac{f”(x)}{2}+\frac{f”(x)}{2} = f”(x)

\end{align*}Letting $f(x)=x^2\sin(1/x)$, $f(0)=0$, then $f$ is continuous and differentiable at $x=0$, and since $f$ is an odd function, $f(h)-f(-h)-2f(0)=0$ for all $h\neq 0$ so that the limit above exists for $x=0$. However $f’$ is not continuous at 0, so $f”(0)$ does not exist.

#### Exercise 12

(By analambanomenos) For $x>0$, $f(x)=x^3$, $f’(x)=3x^2$, $f”(x)=6x$, $f^{(3)}(x)=6$, and for $x<0$, $f(x)=-x^3$, $f’(x)=-3x^2$, $f”(x)=-6x$, $f^{(3)}(x)=-6$.

\begin{align*}

f’(0) &= \lim_{t\rightarrow0}\frac{f(t)}{t}=\lim_{t\rightarrow0}\bigl(\mathop{\rm sign}(t)t^2\bigr)=0 \\

f”(0) &= \lim_{t\rightarrow0}\frac{f’(t)}{t}=\lim_{t\rightarrow0}\bigl(\mathop{\rm sign}(t)3t\bigr)=0

\end{align*} $f^{(3)}(0)$ does not exist for otherwise $f^{(3)}(x)$ would have a simple discontinuity at $x=0$, which cannot happen by the Corollary to Theorem 5.12.

#### Exercise 13

(By analambanomenos) First note that $\sin\bigl(|x|^{-c}\bigr)$ fluctuates between $-1$ and 1, and each neighborhood of 0 has an infinite number of elements from each of the sets $f^{-1}(0)$, $f^{-1}(1)$ and $f^{-1}(-1)$.

(a) $f$ is continuous at 0 if and only if $\lim_{x\rightarrow0}f(x)=0$ which only happens if $\lim_{x\rightarrow0}|x|^a=0$, that is $a>0$.

(b) $f’(0)=\lim_{t\rightarrow0}f(t)/t=\lim_{t\rightarrow0}t^{a-1}\sin\bigl(|t|^{-c}\bigr)$ exists if and only if $\lim_{x\rightarrow0}|x|^{a-1}$ exists, that is $a>1$.

(c) For $x>0$, $$f’(x)=ax^{a-1}\sin(x^{-c})-cx^{a-c-1}\cos(x^{-c})$$ which is bounded on $(0,1]$ if and only if $a-c-1>0$, that is $a>1+c$. By symmetry, the same is true on $[-1,0)$.

(d) $f’$ is continuous at 0 if and only if $\lim_{x\rightarrow0}f’(x)=0$ which only happens if $\lim_{x\rightarrow0}|x|^{a-c-1}=0$, that is $a>1+c$.

(e) $\lim_{t\rightarrow0+}f’(t)/t=\lim_{t\rightarrow0+}\bigl(at^{a-2}\sin(t^{-c})-ct^{a-c-2}\cos(t^{-c})\bigr)$ exists and is equal to 0 if and only if $\lim_{x\rightarrow0}|x|^{a-c-2}$ exists, that is $a>2+c$. We have the same result when taking the limit from the left.

(f) For $x>0$, $$f”(x)=\bigl(a(a-1)x^{a-2}-c^2x^{a-2c-2}\bigr)\sin(x^{-c})-\bigl(acx^{a-c-1}+c(a-c-1)x^{a-c-2}\bigr)\cos(x^{-c})$$ which is bounded on $(0,1]$ if and only if $a-2c-2>0$, that is $a>2+2c$. By symmetry, the same is true on $[-1,0)$.

(g) $f”$ is continuous at 0 if and only if $\lim_{x\rightarrow0}f”(x)=0$ which only happens if $\lim_{x\rightarrow0}|x|^{a-2c-2}=0$, that is $a>2+2c$.

#### Exercise 14

(By analambanomenos) If $f”(x)$ exists for every $x\in(a,b)$, then $f”(x)\ge0$ for all $x\in(a,b)$ if and only if $f’$ is monotonically increasing, so the second part of the Exercise follows from the first part.

Suppose that $f$ is convex and let $a<x<y<b$. By Exercise 4.23 $$\frac{f(t)-f(x)}{t-x}\le\frac{f(u)-f(y)}{u-y}$$ for $t$ and $u$ close to $x$ and $y$, respectively. Taking limits as $t\rightarrow x$ and $u\rightarrow y$, we get $f’(x)\le f’(y)$.

Conversely, suppose is $f’$ is monotonically increasing, let $a<x<y<b$, and let $z=\lambda x+(1-\lambda)y$ for $0<\lambda<1$. By Theorem 5.10 there are points $w_1,w_2$ such that $x<w_1<z<w_2<y$ such that

\begin{align*}

\frac{f(z)-f(x)}{z-x}=f’(w_1) &\le f’(w_2)=\frac{f(y)-f(z)}{y-z} \\

\frac{f(z)-f(x)}{(\lambda-1)x+(1-\lambda)y} &\le \frac{f(y)-f(z)}{\lambda(y-x)} \\

f(z)(y-x) &\le \bigl(\lambda(y-x)\bigr)f(x)+\bigl((1-\lambda)(y-x)\bigr)f(y) \\

f\bigl(\lambda x+(1-\lambda)y\bigr) &\le \lambda f(x)+(1-\lambda)f(y)

\end{align*} which shows that $f$ is convex on $(a,b)$. (The algebra above is much easier if you just take $\lambda=1/2$, then apply Exercise 4.24.)

**Baby Rudin 数学分析原理完整第五章习题解答**