Existence of Derivatives in a Neighborhood

[e(ho0n3]

Simple question: Let f be a real-valued function defined an open interval U. If f is twice differentiable at u, then f' is continuous at u right? Does that mean that there exists a neighborhood around u where f' exists for all points in this neighborhood?

 

[slider142]

That is true, however, keep in mind that f' may not be differentiable in a neighborhood of u (ie., think of the infamous function f(x) = x for rational x, and 0 otherwise, which is only continuous at 0).

 

[e(ho0n3]

What would be the proof of this though? If I use the epsilon-delta definition of the derivative for f' at u, we get the following:

For any e > 0 there is a d > 0 such that |x - u| < d implies |f'(x) - f'(u) - f''(u)(x - u)| < e|x - u|.

This doesn't say that f'(x) exists for x satisfying |x - u| <d because the epsilon-delta definition assumes the function is already defined properly, so as a matter of fact, we can't even employ the epsilon-delta definition.

 

[jostpuur]

Simple question: Let f be a real-valued function defined an open interval U. If f is twice differentiable at u, then f' is continuous at u right?(2) Does that mean that there exists a neighborhood around u where f' exists for all points in this neighborhood?(1)

An answer to 1:

Consider the following question: "If f&#039;(x_0) exists, then does f(x) exist for x_0-\delta &lt; x &lt; x_0+\delta with some delta?"

Do you see why that is confusing question? The derivative is defined by the formula

<br /> f&#039;(x_0) = \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h},<br />

so you cannot define the derivative in the first place, if f doesn't exist in some environment of x_0. In any case, the answer to the question is "yes", because by definition, the f must have existed before the derivative was defined. It is not a kind of thing you prove, instead you read it from the definition.

In your original question you made this one step trickier by assuming that f&#039;&#039;(x_0) exists, and then asked about f&#039;(x) in the environment. But it's the same thing.

An answer to 2:

A derivative never exists in a point of discontinuity.

 

[e(ho0n3]

An answer to 1:

Consider the following question: "If f&#039;(x_0) exists, then does f(x) exist for x_0-\delta &lt; x &lt; x_0+\delta with some delta?"

Do you see why that is confusing question? The derivative is defined by the formula

<br /> f&#039;(x_0) = \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0)}{h},<br />

so you cannot define the derivative in the first place, if f doesn't exist in some environment of x_0. In any case, the answer to the question is "yes", because by definition, the f must have existed before the derivative was defined. It is not a kind of thing you prove, instead you read it from the definition.

In your original question you made this one step trickier by assuming that f&#039;&#039;(x_0) exists, and then asked about f&#039;(x) in the environment. But it's the same thing.

Thanks for the clarification. I think I can also argue by contradiction: if f' was not defined for any neighborhood around u, then f''(u) would not exists by definition.