Proof, arithmetic of uniform continuity

[moses]

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...

 

[tkhunny]

Try sticking a point between them and use the Schwartz Inequality.

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

That may lead somewhere.

 

[moses]

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)|

?

 

[moses]

Try sticking a point between them and use the Schwartz Inequality.

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

That may lead somewhere.

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

 

[tkhunny]

Traingle Inequality?

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

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

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

Or should we be using "M"?