# limits and uniformly continuous

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• Mar 24th 2010, 07:36 PM
inthequestofproofs
limits and uniformly continuous
I need to show that if f is uniformly continuous on (a,b], then the right hand limit of f(x) exists.

This is what i have so far:
If f is uniformly continuous on (a,b], given epsilon >0, there is a delta > 0 s.t. |x-y| < delta -> |f(x)-f(y)| < epsilon for all x,y in (a,b]
Let x be in (a, a + delta).
|f(x) - f(y)|
<= |f(x) - L|+|L-f(y)|
< |f(x) - L|
< epsilon.

Thus, the right hand limit of f exist

Does this make sense?
• Mar 25th 2010, 04:47 AM
Laurent
Quote:

Originally Posted by inthequestofproofs
I need to show that if f is uniformly continuous on (a,b], then the right hand limit of f(x) exists.

This is what i have so far:
If f is uniformly continuous on (a,b], given epsilon >0, there is a delta > 0 s.t. |x-y| < delta -> |f(x)-f(y)| < epsilon for all x,y in (a,b]
Let x be in (a, a + delta).
|f(x) - f(y)|
<= |f(x) - L|+|L-f(y)|
< |f(x) - L|
< epsilon.

Thus, the right hand limit of f exist

Does this make sense?

Not really. First, we don't know what your $\displaystyle L$ is; second, $\displaystyle |f(x)-L|+|L-f(y)|<|f(x)-L|$ is more than dubious.

Since you have to prove the existence of a limit and your function is not monotone, you should use Cauchy condition. This should be pretty straightforward (no computation at all).
• Mar 25th 2010, 07:32 AM
inthequestofproofs
sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?
• Mar 25th 2010, 11:13 AM
Laurent
Quote:

Originally Posted by inthequestofproofs
sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?

You have the same criterion as for sequences.

Fact (Cauchy property for functions) : $\displaystyle f(t)$ has a limit when $\displaystyle t\to a^+$ if, and only if, for every $\displaystyle \epsilon>0$, there is $\displaystyle t_0>a$ such that, for any $\displaystyle t,t'\in(a,t_0)$, $\displaystyle |f(t)-f(t')|<\epsilon$.

Proof:
a) Assume $\displaystyle f(t)\to_{t\to a^+}\ell$. Let $\displaystyle \epsilon>0$. There is $\displaystyle t_0>a$ such that, if $\displaystyle a<t<t_0$, $\displaystyle |f(t)-\ell|<\epsilon/2$. Then, for all $\displaystyle a<t,t'<t_0$, $\displaystyle |f(t)-f(t')|\leq |f(t)-\ell|+|\ell-f(t')|\leq \epsilon$.

b) Assume Cauchy condition : for every $\displaystyle \epsilon>0$, there is $\displaystyle t_0>a$ such that, for any $\displaystyle a<t,t'<t_0$, $\displaystyle |f(t)-f(t')|<\epsilon$. Choose a particular sequence $\displaystyle (u_n)_{n\in\mathbb{N}}$ such that $\displaystyle u_n\to_n a^+$. Then the assumption shows that $\displaystyle (f(u_n))_{n\in\mathbb{N}}$ is a Cauchy sequence (I let you check that). Therefore, it has a limit $\displaystyle \ell\in\mathbb{R}$ : $\displaystyle f(u_n)\to_n \ell$.
Let us show that $\displaystyle f(t)\to_{t\to a^+} \ell$. Let $\displaystyle \epsilon>0$. Let $\displaystyle t_0$ be given by the assumption. There is $\displaystyle n_0$ such that $\displaystyle n\geq n_0$ implies $\displaystyle a<u_n<t_0$ (because $\displaystyle u_n\to_n a^+$) and $\displaystyle |f(u_n)-\ell|<\epsilon$ (because $\displaystyle f(u_n)\to_n\ell$). For every $\displaystyle a<t<t_0$, we have $\displaystyle |f(t)-\ell|\leq |f(t)-f(u_n)|+|f(u_n)-\ell|\leq 2\epsilon$ (first term because of Cauchy condition since $\displaystyle a<u_n,t<t_0$). This concludes.

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In your problem, the uniform continuity immediately gives the Cauchy condition.