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Math Help - continous functions on intervals: existence of abs. min and max

  1. #1
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    continous functions on intervals: existence of abs. min and max

    hello,

    I have 2 problems here:

    1) Let I = [0, pi/2] and let f:I->R be defined as f(x) = sup{x^2, cos x} for x E I. Show that their exists an absolute min point at u E I for f on I. Show that u is the solution to the equation cos x = x^2.

    - attempt at a solution:
    define f(x) = cos x when cos x>= x^2
    = x^2 when x^2 > cos x
    let x be such that f(x) = cos x. thus cos x is continuous for all x in R
    let x be such that f(x) = x^2. x^2 is a polynomial and is continuous for all x in R
    Does this imply that f(x) is continuous?
    assuming that I proved that f(x) is continuous...
    then since I is bounded and f is continuous on I, there exists an absolute min u in I for f on I.

    i know that cos x = 1 @ 0 and x^2 = 0 at 0, so then f(0) = 1.
    cos x decreases to 0, and x^2 increase to pi^2/4 (both monotonically to x = pi/2).
    ....

    2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

    - attempt at a solution:
    since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
    show that f(I) is bounded.
    NO idea

    thanks
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sprinks13 View Post
    hello,

    I have 2 problems here:

    1) Let I = [0, pi/2] and let f:I->R be defined as f(x) = sup{x^2, cos x} for x E I. Show that their exists an absolute min point at u E I for f on I. Show that u is the solution to the equation cos x = x^2.

    - attempt at a solution:
    define f(x) = cos x when cos x>= x^2
    = x^2 when x^2 > cos x
    let x be such that f(x) = cos x. thus cos x is continuous for all x in R
    let x be such that f(x) = x^2. x^2 is a polynomial and is continuous for all x in R
    Does this imply that f(x) is continuous?
    assuming that I proved that f(x) is continuous...
    then since I is bounded and f is continuous on I, there exists an absolute min u in I for f on I.

    i know that cos x = 1 @ 0 and x^2 = 0 at 0, so then f(0) = 1.
    cos x decreases to 0, and x^2 increase to pi^2/4 (both monotonically to x = pi/2).
    ....

    2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

    - attempt at a solution:
    since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
    show that f(I) is bounded.
    NO idea

    thanks
    1. Define h(x)=\cos(x)-x^2. It is apparent that h(x) is continuous, and since h(0)>0 and h\left(\frac{\pi}{3}\right)<0 we can conclude by the Intermediate Value Theorem that there must be zero in \left[0,\frac{\pi}{2}\right]~{\color{red}(*)}.Note that \cos(x) and x^2 are both monotonic on \left[0,\frac{\pi}{2}\right] so there can be only one intersection, which we shall denote \xi. A quick sketch of the graph will then reveal that

    f(x)=\max\left\{\cos(x),x^2\right\}=\left\{ \begin{array}{rcl} \cos(x) & \mbox{if} & 0\leqslant x \leqslant \xi\\<br />
x^2 & \mbox{if} & \xi<x\leqslant\frac{\pi}{2}\end{array}\right.

    So the only possible point of discontinuity is x=\xi. To check this we first notice three things: \xi is a limit point of \left[0,\frac{\pi}{2}\right], that \cos(x), and x^2 are left and right hand continuous respectively. So

    \lim_{x\to\xi^-}f(x)=\lim_{x\to\xi^-}\cos(x)=\cos(\xi)

    And

    \lim_{x\to\xi^+}f(x)=\lim_{x\to\xi^+}x^2=\xi^2

    And finally

    f(\xi)=\cos(\xi)

    Now remembering that \xi is the unique point in \left[0,\frac{\pi}{2}\right] such that x^2=\cos(x) we can conclude that \lim_{x\to\xi^-}f(x)=f(\xi)=\lim_{x\to\xi^+}f(x), thus f(x) is continuous on \left[0,\frac{\pi}{2}\right].

    Now is here where I am not sure what level of math you are at. Tell me if this is beyond the scope of your course

    Theorem: If f:X\mapsto Y is continous and X is compact, then so is f(X)

    Proof: Let \left\{\Psi_n\right\} be an open covering of f(X). Using the fact that the inverse image of any open set in Y must be open in X we can see that \left\{f^{-1}\left(\Psi_{n}\right)\right\} is an opening cover of X. But since X is compact it follows that X\subset f^{-1}\left(\Psi_{n_1}\right)\cup f^{-1}\left(\Psi_{n_2}\right)\cup\cdots. Using the identity f\left(f^{-1}\left(E\right)\right)\subset E we can concude that f\left(X\right)\subset \Psi_{n_1}\cup\Psi_{n_2}\cup\cdots thus f\left(X\right) is compact.

    So knowing by the Heine-Borel Theorem that any closed set [a,b] is compact we can conclude that f\left([a,b]\right) is compact, thus closed and bounded, and due to its boundedness we know that the supremum and infimum are included.

    So back to our actual problem since we have that f:\left[0,\frac{\pi}{2}\right]\mapsto E. So since \left[0,\frac{\pi}{2}\right] is both bounded and closed it is compact, and since f(x) is continuous it follows that so is f\left(\left[0,\frac{\pi}{2}\right]\right)=E, thus \sup_{x\in\left[0,\frac{\pi}{2}\right]}f(x),\inf_{x\in\left[0,\frac{\pi}{2}\right]}f(x)\in E, which is what was asked to be proven.


    I'll come back later for number two.

    Note: \color{red}(*) was gotten by f being continuous and \left[0,\frac{\pi}{2}\right] being connected...in case you are in analysis.
    Last edited by Mathstud28; December 14th 2008 at 08:12 PM. Reason: Typo
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  3. #3
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    Sorry, never talked about something being compact.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sprinks13 View Post
    Sorry, never talked about something being compact.
    Ok, basically in \mathbb{R} compact is synonomous with closed and bounded. So basically my theorem is saying that if f is continuous then f\left([a,b]\right)=E is closed and bounded...does this make sense?
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  5. #5
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    yes, that makes sense. Thanks!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sprinks13 View Post

    2) suppose f:R->R is continuous on R and that lim(x->infinity)f = 0 and lim(x-> - infinity) f = 0. Prove that f is bounded and that it attains either a minimum or a maximum.

    - attempt at a solution:
    since I = R is an interval, and f is continuous on I, then f(I) is also an interval.
    show that f(I) is bounded.
    NO idea

    thanks
    I gave you the last one, so here are some hints

    1. To prove that it there is a maximum or minimum apply Rolle's Theorem
    2. To prove that it is bounded note that unboundedness is caused by either a simple discontinuity such as \lim_{x\to a}f(x)\to\pm\infty or f is unbounded as |x| gets arbitrarily large. The latter is obviously not true and argue that the former is also false based on the continuity of f
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