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Math Help - Question with a Gigantic Hint that makes no sense

  1. #1
    Senior Member slevvio's Avatar
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    Question with a Gigantic Hint that makes no sense

    Hello there, I have this question that comes with a hint; however the hint doesnt seem to make any sense to me whatsoever so I would appreciate some insight into this.

    Let f : \mathbb{R} \rightarrow \mathbb{R} be a function. We define f(x) \rightarrow 0 as x \rightarrow \infty to mean that

    \forall \epsilon > 0 \exists K > 0 such that x > K \implies |f(x)| < \epsilon

    Similarly define f(x) \rightarrow 0 as x \rightarrow - \infty to mean that

    \forall \epsilon > 0 \exists K > 0 such that x < - K \implies |f(x)| < \epsilon

    Now suppose that f: \mathbb{R} \rightarrow [0, \infty ) is continuous and has f(x) \rightarrow 0 as x \rightarrow \infty and f(x) \rightarrow 0 as  x \rightarrow -\infty. Show that f is bounded and attains its maximal value on \mathbb{R} , i.e. there is some v \in \mathbb{R} with f(x) \le f(v) \forall x \in \mathbb{R} .

    [HINT: If f(x) = 0 for all x, there is nothing to prove (why not? [well because f(x)= a constant so it is bounded by that and that is its maximal value for all x] so we can assume that there is some x \in \mathbb{R} with f(x) > 0. Then take \epsilon = \frac{f(x)}{2} and find K_1, K_2 > 0 such that -K_1 < x < K_2 and |f(y)| < \epsilon for y < - K_1 or y > K_2. Now use the extreme value theorem ]

    Solution: or as much as I can do...

    Now  \exists x \in \mathbb{R} such that f(x) > 0.

    Let  \epsilon = \frac{f(x)}{2}. Choose K_1 > 0 such that y<-K_1 \implies |f(y)| < \epsilon = \frac{f(x)}{2}.
    Choose K_2 > 0 such that  y> K_2 \implies |f(y)| < \epsilon = \frac{f(x)}{2} .

    But x< -K_1 \implies |f(x)| = f(x) < \frac{f(x)}{2} which is a contradiction.

    And x> K_2 \implies |f(x)| = f(x) < \frac{f(x)}{2} which is a contradiction.

    Hence -K_1 \le x \le K_2.

    But this doesn't seem to show me anything?? Any help with this would be deeply appreciated.
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  2. #2
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    Simply observe that f attains its maximum (that we know which is \geq f(x)) in the compact interval [K_1,K,2]
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    Quote Originally Posted by slevvio View Post
    Hello there, I have this question that comes with a hint; however the hint doesnt seem to make any sense to me whatsoever so I would appreciate some insight into this.

    Let f : \mathbb{R} \rightarrow \mathbb{R} be a function. We define f(x) \rightarrow 0 as x \rightarrow \infty to mean that

    \forall \epsilon > 0 \exists K > 0 such that x > K \implies |f(x)| < \epsilon

    Similarly define f(x) \rightarrow 0 as x \rightarrow - \infty to mean that

    \forall \epsilon > 0 \exists K > 0 such that x < - K \implies |f(x)| < \epsilon

    Now suppose that f: \mathbb{R} \rightarrow [0, \infty ) is continuous and has f(x) \rightarrow 0 as x \rightarrow \infty and f(x) \rightarrow 0 as  x \rightarrow -\infty. Show that f is bounded and attains its maximal value on \mathbb{R} , i.e. there is some v \in \mathbb{R} with f(x) \le f(v) \forall x \in \mathbb{R} .

    [HINT: If f(x) = 0 for all x, there is nothing to prove (why not? [well because f(x)= a constant so it is bounded by that and that is its maximal value for all x] so we can assume that there is some x \in \mathbb{R} with f(x) > 0. Then take \epsilon = \frac{f(x)}{2} and find K_1, K_2 > 0 such that -K_1 < x < K_2 and |f(y)| < \epsilon for y < - K_1 or y > K_2. Now use the extreme value theorem ]

    Solution: or as much as I can do...

    Now  \exists x \in \mathbb{R} such that f(x) > 0.

    Let  \epsilon = \frac{f(x)}{2}. Choose K_1 > 0 such that y<-K_1 \implies |f(y)| < \epsilon = \frac{f(x)}{2}.
    Choose K_2 > 0 such that  y> K_2 \implies |f(y)| < \epsilon = \frac{f(x)}{2} .

    But x< -K_1 \implies |f(x)| = f(x) < \frac{f(x)}{2} which is a contradiction.

    And x> K_2 \implies |f(x)| = f(x) < \frac{f(x)}{2} which is a contradiction.

    Hence -K_1 \le x \le K_2.

    But this doesn't seem to show me anything?? Any help with this would be deeply appreciated.

    Well I think the hint is a huge one, and in fact it is practically the solution: in general and from what they give you, you have that that |f(y)| \leq e=\frac{f(x)}{2} \,,\,\forall y<-K_1\,\,and\,\,\,\forall y>K_2 , so the function is bounded in (-\infty,-K_1)\cup (K_2,\infty) , and we're left only to worry about boundness of the function in [-K_1,K_2]...but this is a closed bounded interval, and f is continuous, so...

    Tonio
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  4. #4
    Senior Member slevvio's Avatar
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    What does compact mean ?
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    Quote Originally Posted by slevvio View Post
    What does compact mean ?

    In \mathbb{R} this is the same as closed and bounded.

    Tonio
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    Quote Originally Posted by slevvio View Post
    What does compact mean ?
    If you don't know this concept, at least to solve the problem yo must know that in a closed and bounded interval [K_1,K_2] each continuous function attains its maximum. It's the unique thing you have to add to your argument to conclude.
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  7. #7
    Senior Member slevvio's Avatar
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    Thanks for the help I knew I was on the cusp of solving it hehe
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  8. #8
    Senior Member slevvio's Avatar
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    Simple Question about the Extreme Value Theorem

    ok, managed to prove it now thanks
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