Proof of continuous. f(x+y)=f(x)+f(y)

[Unassuming]

Let f be a real-valued function on R satisfying f(x+y)=f(x)+f(y) for all x,y in R.

If f is continuous at some p in R, prove that f is continuous at every point of R.

Proof: Suppose f(x) is continuous at p in R. Let p in R and e>0. Since f(x) is continuous at p we can say that for all e>0, there exists a d>0 such that if 0<|x-p|<d implies that |f(x)-f(p)|< e/2.


I know that |f(x)-f(y)| = |f(x)-f(p)+f(p)-f(y)| <= |f(x)-f(p)|+|f(y)-f(p)|.

I also know that |x-y|=|(x-y+p)-p| <= |x-p|+|y-p|.

Help: I am confused on how to show that f is cont. at every point in R. I appreciate any help little or big! Thank you

 

[mutton]

I also know that |x-y|=|(x-y+p)-p| <= |x-p|+|y-p|.

So if |x - y| < d, then |(x - y + p) + p| < d. Then...

Now use the fact that f(x + y) = f(x) + f(y).

 

[Unassuming]

Let f be a real-valued function on R satisfying f(x+y)=f(x)+f(y) for all x,y in R.

If f is continuous at some p in R, prove that f is continuous at every point of R.

Proof: Suppose f(x) is continuous at p in R. Since f(x) is continuous at p we can say that for all e>0, there exists a d>0 such that if 0<|x-p|<d implies that |f(x)-f(p)|< e.

Let y be a real number and fix 0<|x-y|<d.
Then |(x-y+p)-p|<d
So |f(x-y+p)-f(p)|<e.
or in other words, |f(x-y)+f(p)-f(p)|<e.
or, |f(x)-f(y)|<e.


I have a concern that I used f(x-y)=f(x)-f(y). Does this work because f(-y)= -f(y) ?

I also changed my e/2 into just e.
I also included a "let" before the manipulating.

Does this look better? I appreciate the help mutton.

 

[HallsofIvy]

You don't need to go back to the basic definition of limit. A function, f, is continuous at x= a if and only if:
f(a) is defined
$\lim_{x\rightarrow a} f(x)$ is defined
$\lim_{x\rightarrow a} f(x)= f(a)$

Suppose f is continuous at x= a. Then $\lim_{x\rightarrow a} f(x)= f(a)$

Let h= x- a. Then x= a+ h and as x goes to a, h goes to 0:
$\lim_{x\rightarrow a} f(x)=\lim_{h\rightarrow 0} f(a+ h)= \lim{h\rightarrow 0}f(a)+ f(h)$

What does that tell you about f at x= 0?
(It is important to recognize that if f(x+ y)= f(x)+ f(y), then f(x)= f(x+ 0)= f(x)+ f(0) so f(0)= 0.)

Now let b be any other value and go the opposite way.

 

[mutton]

I have a concern that I used f(x-y)=f(x)-f(y). Does this work because f(-y)= -f(y) ?

It works because f(x) = f(x - y + y) = f(x - y) + f(y).

But what you say is also true because f(-y) = f(0 - y) = f(0) - f(y) = -f(y), as HallsofIvy showed that f(0) = 0.

 

[DDW]

You don't need to go back to the basic definition of limit. A function, f, is continuous at x= a if and only if:
f(a) is defined
$\lim_{x\rightarrow a} f(x)$ is defined
$\lim_{x\rightarrow a} f(x)= f(a)$

Suppose f is continuous at x= a. Then $\lim_{x\rightarrow a} f(x)= f(a)$

Let h= x- a. Then x= a+ h and as x goes to a, h goes to 0:
$\lim_{x\rightarrow a} f(x)=\lim_{h\rightarrow 0} f(a+ h)= \lim{h\rightarrow 0}f(a)+ f(h)$

What does that tell you about f at x= 0?
(It is important to recognize that if f(x+ y)= f(x)+ f(y), then f(x)= f(x+ 0)= f(x)+ f(0) so f(0)= 0.)

Now let b be any other value and go the opposite way.

I was just wondering what you meant by "let b be any other value and go the opposite way."
do you mean let h=x+b?
Thanks.

 

[HallsofIvy]

Actually, h= x- b. If b is any real number, then then $\lim_{x\to b} f(x)= \lim_{h\to 0}f(b+ x)$$= \lim_{h\to 0}f(b)$$+ \lim_{h\to 0}f(h)$$= f(b)+ \lim_{h\to 0}f(h)= f(b)$.

 

[DDW]

Thanks!

 

[quantchem]

What is the exact difference between the h = x-a and h =x-b step? What is the point of doing both steps?