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Let $f(x)$ be continuous on $(0,1]$. Show that $f$ is uniformly continuous IFF $\displaystyle \lim_{x\to0^+} f(x)$ exists.


Backward Proof:

Let another function $\overline f(x)$ be continuous on $[0,1]$ which is equal to $f(x)$ plus the limit point. Thus the limit exists and it is uniformly continuous. So, as $f(x)$.

Forward Proof: I Don't really have any idea...??

Please help guys

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This is the same question as… – Greg Martin Mar 19 '13 at 15:33
up vote 5 down vote accepted

You want to show that $f$ uniformly continuous on $(0,1]\Longrightarrow$ the $\lim_{x\to0^+}f(x)$ exists.

Here are some hints:

  • Let $(x_n)_{n\in\mathbb N}$ be a sequence on $(0,1]$ such that $x_n\to0$. Show (using the uniform continuity of $f$) that $(f(x_n))_{n\in\mathbb N}$ is Cauchy therefore it is convergent.
  • Now if $(x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}$ are two sequences on $(0,1]$ such that $x_n\to0, \ y_n\to 0$, by considering the sequence $(z_n)_{n\in\mathbb N}$ defined as $z_{2k}=x_{2k}$ and $z_{2k+1}=y_{2k+1} \ \forall k\in\mathbb N$, show that $\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(y_n)$.
    Further hint: The sequences $f(x_n),f(y_n),f(z_n)$ converge and $f(x_n),f(y_n)$ are subsequences of $f(z_n)$.
  • Conclude that the $\lim_{x\to0^+}f(x)$ exists.
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Why do we care about odd and even number $n$ and make up another seqence $y_n$ when you have defined $n\in\mathbb{N}$ on your first hint? – Paul Mar 19 '13 at 14:53
Because it's a way to show that the limit of $f(x_n)$ is independent of the sequence $\{x_n\}$. You can do that step a different way if you prefer. – Greg Martin Mar 19 '13 at 15:34
@Paul: In the first hint we show that if $x_n\to0$ then the sequence $f(x_n)$ converges. In the second hint we show that the $\lim f(x_n)$ is independed of the sequence $(x_n)$, i.e. if $x_n\to0$, $y_n\to0$ then $\lim f(x_n)=\lim f(y_n)$. – P.. Mar 19 '13 at 19:10

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