1

The following text is from the book Advanced Calculus by Patrick M. Fitzpatrick :

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There are two things that I don't understand with the proof :

1- In order that the last equation (last line) to hold, the proof must also show that $f(x)$ is continuous but it doesn't. The sequence ${\{f_n}\}$ to be pointwise convergent to $f$ doesn't guarantee that $f$ is contentious on $I$. How to show that $f$ is contentious on $I$?

2- A part of the first claim of the two-claim of the theorem is $f$ is differentiable but it doesn't prove that. How to show that $f$ is differentiable on $I$?

2 Answers2

1

1- $g$ is a uniform limit of continuous functions, therefore it is continuous.

2- $f$ is the integral of a continuous function (see point 1), so it is differentiable.

Giuseppe Negro
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  • I know that $g$ is continuous but my question is to prove continuity of $f$ not $g$. And, what is the proof to what you say at "2"? Thank you –  Feb 20 '16 at 12:03
  • Being differentiable, $f$ is also continuous. The proof of point 2 is a standard theorem in calculus (sometimes called "fundamental formula of calculus"). – Giuseppe Negro Feb 20 '16 at 12:09
  • From http://math.stackexchange.com/questions/429769/is-an-integral-always-continuous I could see that $f(x)-f(x_0)$ is continuous. And for the second question, it is inside The Second Fundamental Theorem. I got it! –  Feb 20 '16 at 12:30
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1

The simplest way to show that the function f is continuous is to show that the function has a two-sided limit for each argument (all x) in the interval I.

Aka you have to prove "$ \forall x \in I: ~ \lim_{{x} \to {x_{x_{0}}}} f(x) = f(x_{0})\\ $" for the funktion.

2

The simplest way to show that the function f is differentiable is to show that the difference quotients of the function has a two-sided limit for each argument (all x) in the interval I.

$ \forall x \in I: ~ \lim_{x \rightarrow x_0} \frac {f(x) - f(x_0)} {x - x_0} = \lim_{h \rightarrow 0} \frac {f(x + h) - f(x)} {h} $

You also would be able to prove it with the integrable of a function:

Are there still questions?

Kevin Dietrich
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