Real analysis show f is continuous

In summary, the function f is continuous at 0, but not everywhere. To show that it is continuous everywhere, we use an epsilon-delta argument to prove that f is continuous everywhere.
  • #1
zachsdado
8
0

Homework Statement


suppose f is a function with the property f(x+y)=f(x)+f(y) for x,y in th reals. suppose f is continuous at 0. show f is continuous everywhere.


Homework Equations





The Attempt at a Solution



I do not understand how to show that f is continuous everywhere.
 
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  • #2
Try to see what properties f has. For example, try to prove that f(x-y) = f(x)-f(y). (Hint: Prove first that f(0)=0 and f(-x)=-f(x).)

Then use an epsilon-delta argument to prove that f is continuous everywhere.
 
  • #3
Use the definition of continuous: f is continuous if and only if [itex]\lim_{x\rightarrow a}f(x)= f(a)[/itex].

Now, let h= x- a and write that in terms of [itex]\lim_{h\rightarrow 0}[/itex].

Exactly the same proof can be used to show that if such an f is continuous at any specific x, then it is continuous at x=0. Putting the two together, if a function satisfying f(x+y)= f(x)+ f(y) is continuous for any x, then it is contnuous for all x.

And, by the way, f must be of the form f(x)= Cx for some number C. The really remarkable thing is that there exist function satisfying f(x+y)= f(x)+ f(y) that are not continuous!
 
  • #4
ok I have deterimed that i can not do this problem. I do not not even understand how two deduce that f(0)=0
 
  • #5
You know that f(x+y)=f(x)+f(y). What happens if you plug in x=y=0?
 
  • #6
you have f(0)=f(0)+f(0)
 
  • #7
So...
 
  • #8
how is this 0
 
  • #9
f(0)-f(0)=f(0)+f(0)-f(0)
0=f(0)
 
  • #10
That's good. Now can you show that f(-x)=-f(x)? (Try to plug in "clever" values for x and y in f(x+y)=f(x)+f(y).)
 
  • #11
If you prefer, I also think an [itex]\epsilon-\delta[/itex] proof is not very hard in this case.
 
  • #12
In fact we can solve this functional equation. Suppose that f is continuous at 0. Then f(0+0)=f(0)+f(0) so f(0)=0. Now [tex]\lim_{y\to 0}f(x+y) = \lim_{y\to 0}f(x)+f(y) = f(x)+f(0)=f(y)[/tex] so [tex]f[/tex] is continuous everywhere. Let [tex]f(1)=c[/tex]. But using the argument f(1+...+1) = nf(1) you can show that f(n)=nc. Now consider a rational number p/q with p,q>0 so if x=p/q then qx=p and so f(qx) = p so f(x+...+x)=p thus pf(x)=ap; that means f(p/q) = a*(p/q). We can extend this argument to negative also easily. And so if x is any real number there is a sequence of rationals converging to it so f(x) = a*x because this is true for rationals. Thus, the only possible functions are the lines through the origin.
 
  • #13
There do, however, exist non-continuous solutions to that functional equation. In distinction to the continuous solutions, all of the form f(x)= Cx, non-continuous solutions are remarkably messy- there are "dense in the plane".
 
  • #14
HallsofIvy said:
There do, however, exist non-continuous solutions to that functional equation. In distinction to the continuous solutions, all of the form f(x)= Cx, non-continuous solutions are remarkably messy- there are "dense in the plane".
Trivial? Use axiom of choice and contruct the Hamel Basis.
 
  • #15
I didn't say it wasn't easy! I'm not sure I would consider contructing a Hamel basis trivial, though. This is one of those situations where it is easy to tell HOW to do something, just very difficult to do it. "First construct a basis for the real numbers as a vector space over the rational numbers"- It's very easy to see that we can use "1" as a basis element and then [itex]\sqrt{2}[/itex] as another, ... but not clear to me how I could give the entire basis. Now define f(1)= 1, f([itex]\sqrt{2}[/itex])= 2, f(v)= 0 for v any other member of the basis, and define f(x) for any real number x "by linearity". Then we have a function satisfying f(x+ y)= f(x)+ f(y) (and f(rx)= rf(x) for any rational r) but which is not continuous.

(I think we have gotten well off the original question now!)
 
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1. What is real analysis and why is it important?

Real analysis is a branch of mathematics that deals with the study of real numbers and their properties. It is important because it provides a rigorous foundation for many other areas of mathematics and is essential for understanding concepts such as continuity and differentiability.

2. What does it mean for a function to be continuous?

A function is continuous if it can be drawn without lifting the pen from the paper. In other words, this means that the function has no jumps, breaks, or holes in its graph.

3. How do you prove that a function is continuous?

To prove that a function is continuous, you must show that it satisfies the epsilon-delta definition of continuity. This means that for any given epsilon (a small positive number), there exists a delta (another small positive number) such that if the distance between any input and the point at which the function is evaluated is less than delta, then the distance between the output and the value of the function at that point will be less than epsilon.

4. Can a function be continuous at some points but not others?

Yes, a function can be continuous at some points but not others. This is known as pointwise continuity. A function is considered continuous if it is continuous at every point in its domain.

5. How is continuity different from differentiability?

Continuity and differentiability are closely related concepts, but they are not the same. A function is continuous if it can be drawn without lifting the pen from the paper, while a function is differentiable if it has a well-defined derivative at every point in its domain. In other words, differentiability is a stronger condition than continuity.

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