Where is the following function continuous

In summary, to show that the function f is continuous for y ≠ 0 and discontinuous for y = 0, we can use the dominated convergence theorem to show that for any sequence y_n approaching y, the limit of f(y_n) is equal to the integral of a function g which is bounded by |y_n\arctan x|. This shows that f is continuous for y ≠ 0. For y = 0, we can use the same approach by choosing a sequence y_n = 0 and showing that the limit of f(y_n) is not equal to the integral of g, thus showing that f is discontinuous at y = 0.
  • #1
firenze
5
0

Homework Statement


[tex]f: [0,+\infty) \to \mathbb{R}: y \mapsto \int_0^{+\infty} y \arctan x \exp(-xy)\,dx.
[/tex]
Show that this function is continuous in [tex]y[/tex] if [tex]y \neq 0[/tex]
and discontinuous if [tex]y = 0[/tex]

Homework Equations





The Attempt at a Solution


I just can't get started, any hint?
 
Physics news on Phys.org
  • #2
Start by trying to simplify an expression for f(y+d)-f(y). Ultimately you want to show that for any epsilon>0, you can pick a delta so that |f(y+d)-f(y)|<epsilon for all d<delta.
 
  • #3
Here is my try:
Choose a sequence [tex]y_n \in [0,+\infty )[/tex] such that [tex]y_n \to y (\neq 0)[/tex].
Define the function [tex]g_n(x)=y_n \arctan x e^{-xy_n}[/tex], then its limit is [tex]g(x)=y\arctan x e^{-xy}[/tex].
Note that [tex]|g_n(x)| \leq |y_n\arctan x|[/tex], it follows [tex]g_n[/tex] is integrable. Hence by dominated convergence thm we have
[tex]\lim f(y_n)=\lim \int g_n \to \int g = f(y)[/tex].

Am I right? Still no idea for the case [tex]y=0[/tex]
 
Last edited:
  • #4
To use the dominated convergence theorem you need to find a function that bounds g_n for all n. In other words, this function can't have an n in it. Other than that you seem to be on the right track. The same idea should work for y=0.
 

1. What is continuity?

Continuity refers to the property of a function or a graph where there are no abrupt changes or breaks. In simpler terms, a continuous function can be drawn without lifting your pencil or pen from the paper.

2. How can you determine if a function is continuous?

A function is continuous if it satisfies the three conditions of continuity: existence, finiteness, and continuity at a point. This means that the function must exist at a given point, have a finite value, and have a limit that equals the function's value at that point.

3. What are the different types of discontinuity?

There are three types of discontinuity: removable, jump, and infinite. A removable discontinuity occurs when there is a hole or gap in the graph that can be filled by redefining the function at that point. A jump discontinuity happens when there is a sudden change in the function's value at a point. An infinite discontinuity occurs when the function approaches either positive or negative infinity at a certain point.

4. How can you use the limit concept to determine continuity?

The limit concept states that if the limit of a function exists at a certain point, then the function is continuous at that point. This means that the function must have a value at that point, and the value of the limit must be equal to the function's value at that point.

5. Can a function be continuous at some points and discontinuous at others?

Yes, a function can be continuous at some points and discontinuous at others. This is known as a piecewise continuous function. It means that the function is defined by different rules in different intervals, and it may or may not be continuous at the endpoints of those intervals.

Similar threads

Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
389
  • Calculus and Beyond Homework Help
Replies
26
Views
897
  • Calculus and Beyond Homework Help
Replies
3
Views
415
  • Math POTW for University Students
Replies
0
Views
214
  • Calculus and Beyond Homework Help
Replies
3
Views
813
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
523
Back
Top