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Thread: a function that has an asymptote is uniformly continuous

  1. #1
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    a function that has an asymptote is uniformly continuous

    Let $\displaystyle f:[0,\infty]\longrightarrow \mathbb{R}$ be a continuous function and $\displaystyle \lim_{x\rightarrow\infty}(f(x)-(ax+b))=0$. Show that $\displaystyle f$ is uniformly continuous.

    I'd like to use this charactetization of uniform continuity. A function $\displaystyle f$ is uniformly continuous iff for all sequences $\displaystyle x_n,\,y_n$ if $\displaystyle |x_n-y_n|\rightarrow 0$, then also $\displaystyle |f(x_n)-f(y_n)|\rightarrow 0$.

    Suppose $\displaystyle |x_n-y_n|\rightarrow 0$.
    If $\displaystyle x_n$ is a bounded sequence, then $\displaystyle y_n$ must be bounded too, and so if $\displaystyle |x_n-y_n|\rightarrow 0$, then also $\displaystyle |f(x_n)-f(y_n)|\rightarrow 0$, because $\displaystyle f$ is continuous and therefore uniformly continuous on a compact set. If, on the other hand, $\displaystyle x_n\rightarrow\infty$, then also $\displaystyle y_n\rightarrow\infty$ and we have

    $\displaystyle |f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0$

    But what if $\displaystyle x_n$ is neither bounded nor convergent to infinity?
    Last edited by ymar; Jun 28th 2011 at 11:01 AM.
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  2. #2
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    Re: a function that has an asymptote is uniformly continuous

    Quote Originally Posted by ymar View Post
    Let $\displaystyle f:[0,\infty]\longrightarrow \mathbb{R}$ be a continuous function and $\displaystyle \lim_{x\rightarrow\infty}(f(x)-(ax+b))=0$. Show that $\displaystyle f$ is uniformly continuous.

    I'd like to use this charactetization of uniform continuity. A function $\displaystyle f$ is uniformly continuous iff for all sequences $\displaystyle x_n,\,y_n$ if $\displaystyle |x_n-y_n|\rightarrow 0$, then also $\displaystyle |f(x_n)-f(y_n)|\rightarrow 0$.

    Suppose $\displaystyle |x_n-y_n|\rightarrow 0$.
    If $\displaystyle x_n$ is a bounded sequence, then $\displaystyle y_n$ must be bounded too, and so if $\displaystyle |x_n-y_n|\rightarrow 0$, then also $\displaystyle |f(x_n)-f(y_n)|\rightarrow 0$, because $\displaystyle f$ is continuous and therefore uniformly continuous on a compact set. If, on the other hand, $\displaystyle x_n\rightarrow\infty$, then also $\displaystyle y_n\rightarrow\infty$ and we have

    $\displaystyle |f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0$

    But what if $\displaystyle x_n$ is neither bounded nor convergent to infinity?
    Let $\displaystyle g=f-(ax+b)$ and $\displaystyle \epsilon>0$. Then there is $\displaystyle M>0$ such that if $\displaystyle x,y\geq M$ then $\displaystyle |g(x)|,|g(y)|\leq\epsilon/3$. Notice also that $\displaystyle f$ is uniformly continuous on $\displaystyle [0,M+\delta_1]$ (by Heine-Cantor), where $\displaystyle \delta_1=\epsilon/3|a|$. So there is $\displaystyle \delta_2>0$ such that if $\displaystyle |x-y|<\delta_2$ for $\displaystyle x,y\in[0,M+\delta_1]$ then $\displaystyle |f(x)-f(y)|<\epsilon$. Now let $\displaystyle \delta=\min\{\delta_1,\delta_2\}$, and suppose $\displaystyle |x-y|<\delta$. If $\displaystyle x,y\geq M$ then $\displaystyle |f(x)-f(y)|<\epsilon$ (details omitted). Otherwise $\displaystyle M>x$ or $\displaystyle M>y$, giving us $\displaystyle x,y\in[0,M+\delta_1]$ and hence $\displaystyle |f(x)-f(y)|<\epsilon$. So in both cases we have $\displaystyle |x-y|<\delta$ implying $\displaystyle |f(x)-f(y)|<\epsilon$.
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