- #1
JG89
- 728
- 1
Homework Statement
Let f and g be continuous functions defined on all of R. Prove that if [tex]f(a) \neq g(x)[/tex] for some [tex] a \epsilon R [/tex], then there is a number [tex] \delta > 0 [/tex] such that [tex] f(x) \neq g(x) [/tex] whenever [tex]|x-a| < \delta[/tex].
Homework Equations
I would like to please check if my proof is correct :)
The Attempt at a Solution
Suppose that [tex] f(x) = g(x) [/tex] whenever [tex] |x-a| < \delta [/tex]. Then by the continuity of f and g we have [tex] |f(x) - f(a)| < \epsilon [/tex] whenever [tex] |x-a| < \delta [/tex] and [tex] |g(x) - g(a)| < \epsilon [/tex] whenever [tex] |x-a| < \delta [/tex]. Since g(x) = f(x) whenever [tex] |x-a| < \delta [/tex], we can write [tex] |g(x) - f(a)| < \epsilon [/tex]. Obviously the left hand side must tend to 0 as x tends to a, and so for values of g(x) > f(a), we have [tex] \lim_{x \rightarrow a} g(x) - \lim_{x \rightarrow a} f(a) = 0 [/tex], implying that [tex] \lim_{x \rightarrow a} g(x) = f(a) [/tex].
Now for values of g(x) < f(x) we have [tex] \lim_{x \rightarrow a} f(a) - \lim_{x \rightarrow a} g(x) = 0 [/tex], implying again that [tex] \lim_{x \rightarrow a} g(x) = f(a) [/tex]. Since g is continuous, we must have [tex] \lim_{x \rightarrow a} g(x) = g(a) [/tex], but limits are unique, so this means that [tex] f(a) = g(a) [/tex], which is a contradiction. QED