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

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

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

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

    And

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

    And finally

    $\displaystyle f(\xi)=\cos(\xi)$

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

    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 $\displaystyle f:X\mapsto Y$ is continous and $\displaystyle X$ is compact, then so is $\displaystyle f(X)$

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

    So knowing by the Heine-Borel Theorem that any closed set $\displaystyle [a,b]$ is compact we can conclude that $\displaystyle f\left([a,b]\right)$ 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 $\displaystyle f:\left[0,\frac{\pi}{2}\right]\mapsto E$. So since $\displaystyle \left[0,\frac{\pi}{2}\right]$ is both bounded and closed it is compact, and since $\displaystyle f(x)$ is continuous it follows that so is $\displaystyle f\left(\left[0,\frac{\pi}{2}\right]\right)=E$, thus $\displaystyle \sup_{x\in\left[0,\frac{\pi}{2}\right]}f(x),\inf_{x\in\left[0,\frac{\pi}{2}\right]}f(x)\in E$, which is what was asked to be proven.


    I'll come back later for number two.

    Note: $\displaystyle \color{red}(*)$ was gotten by $\displaystyle f$ being continuous and $\displaystyle \left[0,\frac{\pi}{2}\right]$ being connected...in case you are in analysis.
    Last edited by Mathstud28; Dec 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 $\displaystyle \mathbb{R}$ compact is synonomous with closed and bounded. So basically my theorem is saying that if $\displaystyle f$ is continuous then $\displaystyle f\left([a,b]\right)=E$ 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 $\displaystyle \lim_{x\to a}f(x)\to\pm\infty$ or $\displaystyle f$ is unbounded as $\displaystyle |x|$ gets arbitrarily large. The latter is obviously not true and argue that the former is also false based on the continuity of $\displaystyle f$
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