Proof, arithmetic of uniform continuity

moses

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Apr 1, 2011
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Hello

I'm trying to prove that if f : A > R and g : A > R are uniformly continuous then f - g if uniformly continuous.

I've tried fiddling with the definition of uniform continuity , i.e. just subtracting the inequalities |f(y) - f(x)| < epsilon and |g(y) - g(x)| < epsilon, but I can't seem to make it add up...
 
Try sticking a point between them and use the Schwartz Inequality.

f(y)p+pf(x)<=f(y)p+f(x)p<2ϵ\displaystyle |f(y) - p + p - f(x)| <= |f(y) - p|+|f(x) - p| < 2\epsilon

That may lead somewhere.
 
I'm not sure I understand the Schwatrz inequality...does it mean that the following inequality is true:

|f(y) - f(x) - g(y) - g(x)| <= |f(y) - f(x)| - |g(y) - g(x)|

?
 
tkhunny said:
Try sticking a point between them and use the Schwartz Inequality.

f(y)p+pf(x)<=f(y)p+f(x)p<2ϵ\displaystyle |f(y) - p + p - f(x)| <= |f(y) - p|+|f(x) - p| < 2\epsilon

That may lead somewhere.

Well I know that f(y)p+pf(x)<ϵ\displaystyle |f(y) - p + p - f(x)| < \epsilon is true, but how can I be sure that f(y)p+f(x)p<2ϵ\displaystyle |f(y) - p|+|f(x) - p| < 2\epsilon ?
 
Traingle Inequality?

No, you don't know that f(y)p+pf(x)<ϵ\displaystyle |f(y) - p + p - f(x)| < \epsilon.

You know f(y)p<ϵ\displaystyle |f(y) - p| < \epsilon and f(x)p<ϵ\displaystyle |f(x) - p| < \epsilon

f(y)p+pf(x)=(f(y)p)+(pf(x))<=f(y)p+pf(x)=f(y)p+f(x)p<ϵ+ϵ=2ϵ\displaystyle |f(y) - p + p - f(x)| = |(f(y)-p)+(p-f(x))| <= |f(y)-p| + |p-f(x)| = |f(y)-p| + |f(x)-p| < \epsilon + \epsilon = 2\epsilon

Or should we be using "M"?
 
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