2

Consider the family $F$ of all $f$ differentiable functions $f: \mathbb R\to\mathbb R$, that satisfy, for any pair of real numbers $x$ and $y$, the condition:

$$\frac{f(x)-f(y)}{x-y} = f'\left(\frac{x+y}{2}\right)$$

Now there is a lot of alternatives t and $f$, and the only true is: "All functions of $F$ are class $C^{\infty}$"

I can not understand why. The function $ax + b$ satisfy the properties (1). But this type of function is not smooth.

(1): $$\frac{f(x)-f(y)}{x-y} = \frac{ax + b - ay - b}{x-y} = a\\f' = a$$

k170
  • 9,045
Gabriela Da Silva
  • 527

1 Answers1

1

The $C^\infty$ function space is defined as the set of functions $f$ being $C^n$ for all $n$. If $f$ has a $n$-th derivative (over its whole definition domain) for any integer $n$, then it is $C^\infty$. If a function has a derivative it is necessarily continuous so talking about continuity of the derivative in this definition is not mandatory. In practice we rather say that $f$ is indefinitely derivable or smooth. Note that the former can have different meanings depending on the context.

For $f(x)=ax+b$ is $C^\infty$, we can compute the $n$-th derivative explicitely for arbitrary large ranks $n$. The first derivative, $n=1$, is $f'(x)=a$ and from there $f^{(n)}(x)=0$ for all $n\ge2$. So it is $C^\infty$.

nicomezi
  • 8,254
  • I was misunderstanding the concept of $C^{\infty}$ so ... thank you. we can see that y->x, the equation reduces to the tautology f'(x) = f'(x). Maybe this is why the answer is all $C^{\infty}$ – Gabriela Da Silva Dec 19 '20 at 06:21
  • No the thing is that $F$ only contains $C^\infty$ functions, but not all $C^\infty$ functions satisfy the equation (consider $\exp$ for example with $x=1$ and $y=0$). The equation gives a link between the derivative at some point and the values of the function. By induction we can deduce the $n$-th derivative for any $n$ and at any point, hence by definition ... The tautology does not allow to deduce anything. – nicomezi Dec 19 '20 at 06:42