Differentiability of $f$ at $a$ implies continuous function Let $f:A \subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a function and $a$ an interior point of $A$. Show that the following properties are equivalent:
$(a)$ $f$ is differentiable in $a$.
$(b)$ There exists a $\lambda \in \mathbb{R}$, a $\delta >0$ and a function $\psi:]- \delta, \delta [ \rightarrow \mathbb{R}$ such that


*

*$]a- \delta ,a+ \delta [ \subseteq A$,

*$\psi(0)=0$ and $\psi$ is continuous in $0$,

*$f(a+h)=f(a)+ \lambda h +h \psi (h)$ for every $h \in ]- \delta , \delta[$.


Could someone give me a hint on how to start on this question? I am a bit lost.
 A: You have to show (a) implies (b) and (b) implies (a).
For (a) implies (b), one good place to start would be to look at the third bullet:


*

*$f(a+h)=f(a)+ \lambda h +h \psi (h)$ for every $h \in ]- \delta , \delta[$.


So you want to prove the existence of this function $\psi$ -- why not just solve for it? We have
$$
\psi(h) = \frac{f(a+h) - f(a) - \lambda h}{h}, \text{ for } h \ne 0
$$
and we know by the second bullet that $\psi(0)$ should be $0$.
So start by defining this to be $\psi$:
$$
\psi(h) := \begin{cases} \frac{f(a+h) - f(a) - \lambda h}{h}, &\text{for } h \ne 0 \\ 0 &\text{for } h = 0\end{cases}
$$
Now, what you have to do is pick $\delta$, and prove that the three bullets are in fact true.
For (b) implies (a), now we are assuming the three bullet points and we have to try to prove $f$ is differentiable at $a$. I would suggest again looking at the third bullet, rewritten in the form we had before,
$$
\psi(h) = \frac{f(a+h) - f(a) - \lambda h}{h}, \text{ for } h \ne 0
$$
Take the limit as $h \to 0$ and see what happens.
