Is a constant as a function continuously differentiable?
Well, it may seem trivial, but I cannot find it on google. Is a constant function continuously differentiable, of all orders?
Thank you.
$\endgroup$2 Answers
$\begingroup$Yes. $f'$ and all higher derivatives are identically equal to zero.
$\endgroup$ $\begingroup$Let's assume this is a constant function on $\mathbb{R}$ (i.e. $f : \mathbb{R} \to \mathbb{R}$, $f(x) = c$ for some fixed $c$, for all $x \in \mathbb{R}$).
Fix any $x_0$. What is
$$ \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h} \ \ ? $$
In particular, are there any points where this limit fails to exist? And at the points where the limit does exist, what is the limit equal to?
No, there are no points where the limit does not exist, and at every point, the limit is equal to $0$. To see this, note that $f(x_0 + h) = c$ and $f(x_0) = c$, so the numerator is always $0$.
Thus, if we define a function $$ f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}, $$ then $f'$ is defined for all $x$, and it is continuous everywhere, as required.
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