Solutions

a) If $f$ is Lipschitz of order $\alpha>0$ at $x$, then

$$|f(x)-f(y)|\le C|x-y|^\alpha \implies -|C||x-y|^\alpha \le f(x)-f(y)\le |C||x-y|^\alpha.$$

But

$$\lim_{y\to x}-|C||x-y|^\alpha = \lim_{y\to x}|C||x-y|^\alpha = 0$$

so

$$\lim_{y\to x}f(x)-f(y)=0$$

and therefore $f$ is continuous at $x$.

b) Let $f$ be Lipschitz of order $\alpha>0$ on the interval $A$. Pick any $\varepsilon>0$. Let

$$\delta=\sqrt[\alpha]{\frac{\varepsilon}{C}}.$$

Then, for any $x,y\in A$ we have

$$|x-y|<\delta=\sqrt[\alpha]{\frac{\varepsilon}{C}} \implies |f(x)-f(y)|\le C|x-y|^\alpha<\varepsilon.$$

c) Let $f$ be differentiable at $x$. Then for any $\varepsilon>0$ there exists $\delta>0$ such that

$$\forall y,\ 0<|y-x|<\delta \implies \left|\frac{f(y)-f(x)}{y-x}-f'(x)\right|<\varepsilon.$$

Hence,

$$\begin{align*} \left|\frac{f(y)-f(x)}{y-x}\right| - |f'(x)| &\le \left|\frac{f(y)-f(x)}{y-x}-f'(x)\right|<\varepsilon \\ &\implies \left|\frac{f(y)-f(x)}{y-x}\right|<\varepsilon+|f'(x)| \\ &\implies |f(y)-f(x)|<(\varepsilon+|f'(x)|)|y-x|. \end{align*}$$

Hence $f$ is Lipschitz of order $1$ at $x$ with $C=\varepsilon+|f'(x)|$. The converse is not necessarily true if we take $f(x)=|x|$ and consider $x=0$. Clearly $f$ is Lipschitz of order $1$ at $x=0$ since

$$\forall y,\quad |0|-|y|=|y|\le |y|=|0-y|,$$

where implicitly $C=1$. However, we know that $f$ is not differentiable at $x=0$ because

$$\lim_{x\to 0^-}f'(x)=-1\ne 1=\lim_{x\to 0^+}f'(x).$$

d) Not necessarily. Consider

$$f(x)=x^{3/2}\sin\left(\frac{1}{x}\right)$$

on the interval $[0,1]$ with $f(0)=0$. Notice that for $x\in(0,1]$,

$$f'(x)=\frac{3}{2}x^{1/2}\sin\left(\frac{1}{x}\right)-x^{-1/2}\cos\left(\frac{1}{x}\right)$$

and

$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h} =\lim_{h\to 0}\frac{h^{3/2}\sin(1/h)}{h} =\lim_{h\to 0}h^{1/2}\sin\left(\frac{1}{h}\right)=0.$$

Hence, $f$ is differentiable on $[0,1]$. However, $f'$ is unbounded. For contradiction, suppose now that $f$ is Lipschitz of order $1$ on $[0,1]$. This means that there is a constant $C$ such that

$$|f(x)-f(y)|\le C|x-y| \implies \left|\frac{f(x)-f(y)}{x-y}\right|\le C$$

for all $x,y\in[0,1]$. But then,

$$-C\le \lim_{x\to y}\frac{f(x)-f(y)}{x-y}\le C,$$

and this contradicts the boundedness of $f'$.

e) This is identical to exercise 36 except that there is a constant $C$ in the inequality.