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Let f defined R and continuous at x=0. Please tell me How can i show that f is uniformly continuous?

mrf
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user180805
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    Everything you need and more is at http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – Chris Culter Oct 04 '14 at 06:24

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I'll try to give you a hint, first, so you know $f$ is continuous at $0$, now try to get the value at $f(0)$ \, (what does $f(0 + 0)$ equals to?) then, read the definition of uniformly continuous function, what can you say about $|f(x) - f(y)| $ ? (try to apply your hypothesis)

Answer:

So you know its continuous at $0$, meaning that given some abritrary $\varepsilon > 0$, there exist some $\delta > 0$ such that if $|x| < \delta$ then $f(x) < \varepsilon$.

You want to prove its uniformly continuous that is, for arbitrary $\varepsilon > 0$, if $x,y$ are two numbers such that $|x-y| < \delta$ then $|f(x) - f(y)| < \varepsilon$.

Now, I affirm that for the first $\delta$ (the one for the continuity at $0$) works. First I need to prove that $f(x - y) = f(x) - f(y)$, but before that note that $$f(0) = f(0 + 0) = f(0) + f(0) $$ So $f(0) =0$, and $$0 = f(0) = f(x-x) = f(x) + f(-x)$$ wich means that $f(-x) = -f(x)$, now I'm ready to prove the uniform continuity of $f$.

If $|x-y| < \delta$ then $$|f(x) - f(y)| = |f(x) + f(-y)| = |f(x-y)| < \varepsilon$$

aram
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