# Thread: Real Analysis Proof help dealing with continuity

1. ## Real Analysis Proof help dealing with continuity

If R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and 0 is in C(f), then f is continuous.

show me a great proof...

2. Originally Posted by antoniowilliams46
If R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and 0 is in C(f), then f is continuous.

show me a great proof...

A great one huh?

This is obvious since $f(0)=f(0)+f(0)\implies f(0)=0$ and thus $0=f(0)=f(x-x)=f(x)+f(-x)\implies f(-x)=-f(x)$. Thus, $|f(x)-f(y)|=|f(x-y)|$ now there exists some $\delta>0$ such that $|z|<\delta\implies |f(z)|<\varepsilon$. Make $|x-y|<\delta$

3. thats all that i need???? nothing more???

4. Originally Posted by antoniowilliams46
thats all that i need???? nothing more???
You need to say more but that is the entire idea.

5. still need help on what to say in the proof

6. i still need a proof for this one

7. Do you understand Drexel proof? If you tell me where you get lost, I may be able to help you.

8. I dont know where to go after abs val of x-y is less than delta.

9. Let z=x-y. Then $|z|=|x-y|<\delta\implies |f(z)| = |f(x-y)| = |f(x)-f(y)|<\varepsilon$.

Another way to see it is: f is continuous at x iff f(x+h)=f(x)+o(1) as $h\to0$. So in this case, f(x+h) = f(x) + f(h) by linearity and f(h)=o(1) by continuity at 0.