# bounded continuous function

• May 20th 2010, 08:46 AM
cp05
bounded continuous function
must a bounded continuous function on R be uniformly continuous?
I know that if a function is continuous on a closed and bdd set then its uniformly continuous, but this says nothing about the set, just the function. Would this be T or False? How would I go about proviing it?

Also, if f and g are uniformly continuous maps of R to R, must the product f*g be uniformly continuous? What if f and g are bounded.
I think this one is true...since you're just multiplying two continuous functions together...but I don't know how to prove this either :(

Thanks.
• May 20th 2010, 08:57 AM
Drexel28
Quote:

Originally Posted by cp05
must a bounded continuous function on R be uniformly continuous?
I know that if a function is continuous on a closed and bdd set then its uniformly continuous, but this says nothing about the set, just the function. Would this be T or False? How would I go about proviing it?

Well, for $(0,1)$ what about $f(x)=\sin\left(\frac{1}{x}\right)$? That is bounded and continuous but not uniformly continuous. Can you see how to generalize?

Quote:

Also, if f and g are uniformly continuous maps of R to R, must the product f*g be uniformly continuous? What if f and g are bounded.
I think this one is true...since you're just multiplying two continuous functions together...but I don't know how to prove this either :(

Thanks.
It isn't true, take $f(x)=g(x)=x$
• May 20th 2010, 09:14 AM
cp05
why is http://www.mathhelpforum.com/math-he...136d2bbd-1.gif not uniformly continuous? aren't all the y-values relatively close together? I don't really understand how to prove or disprove uniform continuity other than saying when the x values are really close together so are the y values...

for the second part, what about when f and g are bounded? i know when the derivative is bounded they are uniform continuous because it will be a lipschitz function, but I haven't read anything about f and g being bounded :(
• May 20th 2010, 11:20 AM
Opalg
Quote:

Originally Posted by cp05
for the second part, what about when f and g are bounded? i know when the derivative is bounded they are uniform continuous because it will be a lipschitz function, but I haven't read anything about f and g being bounded :(

If f and g are both uniformly continuous on R, and are both bounded, then fg will be uniformly continuous, because

\begin{aligned}|f(x)g(x)-f(y)g(y)| &= |f(x)\bigl(g(x)-g(y)\bigr) + \bigl(f(x)-f(y)\bigr)g(y)| \\ &\leqslant |f(x)||g(x)-g(y)| + |f(x)-f(y)||g(y)| \end{aligned}

and you can make the right-hand side of that inequality as small as you want, for x and y sufficiently close together, using the given properties of f and g.