1. ## Another Unif. Cont.

Prove or disprove (by counterex.) $\displaystyle f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $\displaystyle f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $\displaystyle m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\displaystyle \epsilon > 0$ set $\displaystyle \delta = ?$. If $\displaystyle x,a \in (0,1)$, then $\displaystyle |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $\displaystyle (x,a) \in (0,1)$ and $\displaystyle |x-a| < \delta$ then $\displaystyle |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.

2. Originally Posted by Math2010
Prove or disprove (by counterex.) $\displaystyle f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $\displaystyle f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $\displaystyle m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\displaystyle \epsilon > 0$ set $\displaystyle \delta = ?$. If $\displaystyle x,a \in (0,1)$, then $\displaystyle |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $\displaystyle (x,a) \in (0,1)$ and $\displaystyle |x-a| < \delta$ then $\displaystyle |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
The function $\displaystyle f(x) = x\log(1/x)$ is not Lipschitz on (0,1), because its derivative becomes infinite as $\displaystyle x\to0$. But $\displaystyle \textstyle\lim_{x\to0}f(x)$ does exist (it is 0), as is $\displaystyle \textstyle\lim_{x\to1}f(x)$. So f extends to a continuous function on the closed interval [0,1] and is therefore uniformly continuous.

3. Originally Posted by Math2010
Prove or disprove (by counterex.) $\displaystyle f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $\displaystyle f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $\displaystyle m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\displaystyle \epsilon > 0$ set $\displaystyle \delta = ?$. If $\displaystyle x,a \in (0,1)$, then $\displaystyle |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $\displaystyle (x,a) \in (0,1)$ and $\displaystyle |x-a| < \delta$ then $\displaystyle |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
The second one's kind of a pain in the ***. But, clearly both $\displaystyle f(x)=\frac{1}{mx+b},g(x)=\cos(x)$ are uniformly continuous on $\displaystyle I=(0,1)$ since $\displaystyle \left|f'(x)\right|\leqslant\left|\frac{m}{b^2}\rig ht|,\left|g(x)\right|\leqslant 1$ and so they are Lipschitz. And, there is a theorem that says that since $\displaystyle f,g$ are uniformly continuous and $\displaystyle I$ is bounded that $\displaystyle f\cdot g$ is uniformly continuous on $\displaystyle I$.

That's the least painful way I can see.