# Thread: Some uniform continuous problems

1. ## Some uniform continuous problems

Discussing the uniform continuity of the following:

1. $f(x)=x^2, \ \ \ x \in (-1,1)$

Claim: f is uniformly continuous on (-1,1)

Proof.

Given $\epsilon > 0$, pick $\delta = \frac { \epsilon }{2}$, then for each $x,y \in (-1,1)$ with $|x-y|< \delta$, we would have $|x^2-y^2|=|x-y||x+y|< \delta |x+y| < \delta |1+1| = \epsilon$

Q.E.D.

2. $f(x)=e^{-x} \ \ \ x \in [0,\infty )$

Claim: f is uniformly continuous.

Proof.

Given $\epsilon > 0$, pick $\delta >0$, then for each $x,y \in [0, \infty)$ with $|x-y|< \delta$, we would have $|e^{-x} - e^{-y}| \leq |e^0-e^0|=|1-1|=0 < \epsilon$

Q.E.D.

3. $f(x)=x^{ \frac {1}{3} } \ \ \ x \in [0, \infty )$

Claim: f is not uniformly continuous.

So far, I know that $f'(x)= \frac {1}{3x^{ \frac {2}{3}}}$, which is not below above as x approaches to 0.

But is it possible to prove this by defintion?

Proof so far.

Pick $\epsilon =1$, then $\forall \delta > 0$ with $|x-y|< \delta$, I need to show that $|x^{ \frac {1}{3}} - y^ { \frac {1}{3} } | < \epsilon$ But I'm stuck in here, any hints?

Thanks!!!

2. Correct on nos. 1 and 2. For 3, split the interval into two subintervals, [0,1] and [1,∞], and ask why the function should be uniformly continuous on both of them.

3. For the first interval of $[0,1]$: f is continuous on [0,1], a compact set, meaning f is also uniformly continuous.

For the second interval of $[1, \infty )$: The derivative is bounded by 0 below and 1/3 above, therefore it is uniformly continuous.

Is this right? So if a function is uniformly continuous on two subinterval, then it is also uniformly continuous on the whole thing?

4. Yes, and yes.

5. I was rewriting my solutions, and I found a problem in my answer to the function of $e^{-x}$

Is $|e^{-x}-e^{-y}| \leq |e^0-e^0|$? I'm thinking now it is not, since y can be something really big, that would bring the distance out.

Should I consider the series $e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k$ and prove this series is uniformly convergence?

Then I use the M test, since $\frac {1}{k!}(-x)^k \leq \frac {2^kx^k}{(k+1)!}$, and $\sum ^ \infty _{k=0} \frac {2^kx^k}{(k+1)!}$ converges by the ratio test, so f is uniformly convergence, is that right? Thanks.

2. $f(x)=e^{-x} \ \ \ x \in [0,\infty )$

Claim: f is uniformly continuous.
I was rewriting my solutions, and I found a problem in my answer to the function of $e^{-x}$

Should I consider the series $e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k$ and prove this series is uniformly convergence?

Then I use the M test, since $\frac {1}{k!}(-x)^k \leq \frac {2^kx^k}{(k+1)!}$, and $\sum ^ \infty _{k=0} \frac {2^kx^k}{(k+1)!}$ converges by the ratio test, so f is uniformly convergence, is that right? Thanks.
You seem to be confusing uniform continuity with uniform convergence. If you want to show that the series $e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k$ is uniformly convergent on the interval [0,∞) then you use the M-test. But the question asks you to show that the function $e^{-x}$ is uniformly continuous on the interval [0,∞). The easiest way to do that is to observe that the derivative $-e^{-x}$ is bounded on that interval.