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# Prove some continuity properties of a function

Given a function and a closed interval such that for all .

1. Prove is continuous at every point .
2. If is integrable on prove 3. If is any point prove 1. Proof. Let be any point in . Then, for any let . Then Thus, for any , we have whenever . Hence, Therefore, is continuous at every point 2. Proof. Since is a constant (the value of the function evaluated at the constant ), we have Therefore, we can compute, The final inequality follows since, using the monotone property of the integral, Then, since by our hypothesis on , we have 3. Proof. The proof proceeds similarly to that of part (b), Then, since , we break the integral into two pieces (to deal with the fact that for and for ): But now, we know implies , so we have the inequality, 1. Kalle says:

Since I had trouble to understand the last step in part c, I share my approach to it:  • Anonymous says:

Thx!

2. Kalle says:

I think you have a mistake in part (c): in the last inequality you have

c(c-(b+a)) <= -ab.

For example for a=2, b=5 and c=1 we get
c(c-(b+a)) = -6
-ab = -10.

• Kalle says:

ah sorry, my example is wrong since c<a

3. You’ve got a sign error in part (c). It should be -c(b-c) in the 3rd line of the 2nd section. It changes a little how you get back to the end inequality.

• Rori says:

Awesome, thanks! Working on a fix now… you’re right though, it cascades into the rest of the proof.

• Rori says:

I think it should be fixed now. Thanks for letting me know.