On derivatives of higher order

[jwqwerty]

let's assume that f(t) is a real function on [a,b], n is a positive integer
(n-1)th derivative of f is continuous on [a,b], (n)th derivative exists for all t in (a,b)

1. Then can we say that (n-2)th derivative of f is continuous on [a,b]?
2. (n-2)th derivative of f is defined on a and b?

 

[statdad]

If a function g has a derivative g', what do you know about the continuity of g?

 

[jwqwerty]

If a function g has a derivative g', what do you know about the continuity of g?

if g is differentiable on (a,b), it is continuous on (a,b)
thus in the same manner we can say that (n-2)th derivative of f is continuous on (a,b)
but what i want to know is whether (n-2) derivative of f is continuous on a and b

 

[mathman]

You seem to have missed the point of statdad comment. Let g = (n-2) derivative of f. Your assumption is that g' exists and is continuous. Therefore g must be continuous.

 

[HallsofIvy]

Saying that f is differentiable on an interval does NOT mean the derivative is continuous on that interval. For example $f(x)= x^2 sin(1/x)$ if $x\ne 0$, f(0)= 0, then it is easy to show that f(x) is continuous for all x. As long as x is not 0, that deriviative is $f'(x)= 2xsin(1/x)+ x^2(-1/x^2)cos(1/x)= 2xsin(1/x)- cos(1/x)$. If x= 0, we get the derivative from the formula: $\lim_{h\to 0}\frac{h^2 sin(1/h)}{h}= 0$. So the derivative exists for all x but $\lim_{x\to 0} f'(x)= \lim_{x\to 0} 2xsin(1/x)- cos(1/x)$ which does not exist so derivative exists but is not continuous at x= 0. Since the derivative is not necessarily continuous it is not necessarily differentiable so the second derivative does not necessarily exist.

(One can show, using the mean value theorem, that the derivative, while not necessarily continuous, must satisfy the "intermediate value property" so at a point where the limit exists but is not continuous it must be because the limit $\lim_{x\to a}f'(x)$ must not exist.

 

[jwqwerty]

if nth derivative of f at t (where t is in (a,b)) is defined, then this means that (n-1)th derivative of f is differentiable at t.
since (n-1)th derivative of f is differentiable at t, (n-1)th derivative of f is continuous at t.
In the same manner, we can argue that (n-2)th, (n-3)th... 1st derivative of f is differentiable at t and continuous at t.

but what i want to know is whether 1st, 2nd...(n-2)th derivative is defined on endpoints a and b
given that f is defined on [a.b] and (n-1)th derivative of f is continuous on [a,b]