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Math Help - Uniform continuity question

  1. #1
    Lord of certain Rings
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    Uniform continuity question

    A few of my friends had enrolled for an Analysis-I course and had recently written their final exams. One of the questions they were asked to prove was(I am writing it from memory, so forgive me if its meaningless):

    If f is uniformly continuous over \mathbb{R}, then prove that there exists constants a and b such that |f(x)| \leq a|x|+b, \forall x \in \mathbb{R}

    I had not enrolled for the course, but I know the definition of uniform continuity and a couple of simple theorems about it. This problem looked too harsh on uniformly continuous functions and I thought it was a misprint ......But it was not

    Any ideas on how to solve it?
    Do I need anything other than the definition? if yes then please tell me the proof. Else give me a hint.


    Thank you,
    Srikanth
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  2. #2
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    Definition of uniform continuity: \forall\epsilon>0\;\exists\delta>0 such that |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<\varepsilon. Apply this with \varepsilon=1 to get a \delta>0 with |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<1. In particular, with y=0, the triangle inequality gives |x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|.

    Suppose in addition that |x|>\delta/2. Then \delta/2<|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|< \tfrac2{\delta}|x| + |f(0)|.

    Next (assuming from now on that \delta/2<|x|<\delta), |2x-x|<\delta and so |f(2x)-f(x)|<1. By the triangle inequality again, |f(2x)|<2+|f(0)|<\tfrac2{\delta}|2x| + |f(0)|.

    That should suggest an inductive argument: |f(nx)|<\tfrac2{\delta}|nx| + |f(0)|.

    But multiples of the interval [-\delta/2,\delta] cover all the positive reals except for the interval [0,\delta/2]. So take a=\tfrac2{\delta} and b=1+|f(0)|, and you'll have |f(x)|\leqslant a|x|+b for all real x.
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  3. #3
    Moo
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    Hi !
    Quote Originally Posted by Opalg View Post
    |f(x)-f(y)|<1. In particular, with y=0, the triangle inequality gives |x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|.
    That's the part that stuck me
    How can the triangle inequality help ?
    Doesn't it say ||x|-|y|| \leq \dots ?
    So since it's \leq, we cannot know if the "right parts" will still be < to 1 ?
    Can you enlighten me pleaaaase ?
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  4. #4
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    Quote Originally Posted by Moo View Post
    How can the triangle inequality help ?
    Doesn't it say ||x|-|y|| \leq \dots ?
    So since it's \leq, we cannot know if the "right parts" will still be < to 1 ?
    If |a-b|<1 then |a| = |(a-b)+b|\leqslant|a-b|+|b|<1+|b|.
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  5. #5
    Lord of certain Rings
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    Quote Originally Posted by Opalg View Post
    Definition of uniform continuity: \forall\epsilon>0\;\exists\delta>0 such that |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<\varepsilon. Apply this with \varepsilon=1 to get a \delta>0 with |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<1. In particular, with y=0, the triangle inequality gives |x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|.

    Suppose in addition that |x|>\delta/2. Then \delta/2<|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|< \tfrac2{\delta}|x| + |f(0)|.

    Next (assuming from now on that \delta/2<|x|<\delta), |2x-x|<\delta and so |f(2x)-f(x)|<1. By the triangle inequality again, |f(2x)|<2+|f(0)|<\tfrac2{\delta}|2x| + |f(0)|.

    That should suggest an inductive argument: |f(nx)|<\tfrac2{\delta}|nx| + |f(0)|.

    But multiples of the interval [-\delta/2,\delta] cover all the positive reals except for the interval [0,\delta/2]. So take a=\tfrac2{\delta} and b=1+|f(0)|, and you'll have |f(x)|\leqslant a|x|+b for all real x.

    Thanks Opalg. The proof is wonderful since does not use any "theorems". The tricks you have used are fantastic...
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