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Homework Help: L-normed vector space

  1. Feb 8, 2006 #1
    Hi

    I'm given the following assignment which deals with to looks like an L-normed vectorspace:

    Prove that,

    [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

    To prove this do I approach the above as a triangle inequality or as a cauchy-swartz inequality?

    Best Regards,

    Fred
     
  2. jcsd
  3. Feb 8, 2006 #2

    benorin

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    What do you know about f ? How is the norm defined?
     
  4. Feb 8, 2006 #3
    Hello and thank You for Your reply,

    f is defined as follows:

    [tex]f: \mathbb{R}^n \rightarrow \mathbb{R}[/tex] and

    [tex]y1, y2 \in \mathbb{R}^n[/tex]

    Best Regards,

    Fred

     
  5. Feb 8, 2006 #4

    benorin

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    More must be given, (or I am just that tired) since

    [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

    does not, in general (under those conditions), hold for arbitrary f:R^n->R, like, say, put [tex]f(y)=2 ||y||[/tex] with y2=0 and y1=y gives

    [tex]|f(y) - f(0)| = 2||y|| \not\leq || y - 0||[/tex]
     
  6. Feb 8, 2006 #5
    Hello again,

    According to my textbook the first step is to show that

    [tex] f(y_2) \leq || y_1 - y_2|| + f(y_1)[/tex]

    The is a sub-problem of a problem which deals distance from a point to a set.

    f is defined as the distance from a point in [tex] \mathbb{R}^n [/tex] to a subset S of [tex]\mathbb{R}^n[/tex]

    and finally I'm suppose to conclude that f is Uniform continuites on [tex]
    \mathbb{R}^n[/tex]

    Any idears?

    Best Regards,

    Fred

     
  7. Feb 8, 2006 #6

    benorin

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    Since f is defined as the distance from a point in [tex] \mathbb{R}^n [/tex] to a subset S of [tex]\mathbb{R}^n[/tex], we should have

    [tex]f_S(y)=\mbox{inf}\left\{ \|x-y\| : x\in S\right\}[/tex] for [tex]y\in\mathbb{R}^n[/tex]

    I put a subscript of S on f since f depends on S, but consider S fixed and supress the subscript from here on.

    [tex] f(y_2) \leq || y_1 - y_2|| + f(y_1)[/tex] is the statement that

    [tex] \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}[/tex],

    how could you prove that?
     
  8. Feb 8, 2006 #7


    By showing that normed distance from x to y1 and from x to y2 are equal? Cause they belong to the same subset?

    Best Regards
    Fred
     
    Last edited: Feb 8, 2006
  9. Feb 8, 2006 #8

    benorin

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    By the triangle inequality, for all [tex]x,y_1,y_2\in\mathbb{R}^n[/tex], we have

    [tex]\|x-y_2\| \leq \| y_1 - y_2\| +\|x-y_1\|[/tex],

    and since [tex]x\in S\mbox{ and }S\subset\mathbb{R}^n[/tex], we may restrict x to be in S and inf over x in S to obtain

    [tex] \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}[/tex]
     
  10. Feb 8, 2006 #9
    Hello again and thank You for Your answer,

    And this shows, that

    [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

    is true?

    Best Regards

    Fred

     
    Last edited: Feb 8, 2006
  11. Feb 8, 2006 #10

    benorin

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    Swap y1 and y2, see what you get.
     
  12. Feb 8, 2006 #11
    Hello again,


    then

    [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

    since

    [tex] \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} - \| y_1 - y_2\| \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} [/tex] ?

    Best Regards

    Fred

     
  13. Feb 8, 2006 #12

    benorin

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    Of this

    [tex] \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}[/tex]

    you need only first and last bits (including the <= sign), namely

    [tex] \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}[/tex]

    in this, swap y1 and y2, to get

    [tex] \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_2 - y_1\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\}[/tex]

    and since [tex] \| y_2 - y_1\| = \| y_1 - y_2\| ,[/tex] we have this pair of inequalities

    [tex]\boxed{\begin{array}{cc} \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \\ \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \end{array} } [/tex]

    thinking of theses as being of the form

    [tex]A\leq k+B[/tex]
    [tex]B\leq k+A[/tex]

    we then have

    [tex]A-B\leq k[/tex] and
    [tex]B-A\leq k[/tex]

    thus [tex]|B-A| \leq k[/tex] i.e. [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]
     
  14. Feb 8, 2006 #13
    Hello again, and thank You for Your answers,

    I need to conclude that the function f is is uniformly continius on [tex]\mathbb{R}^n[/tex].

    In order to show this by proving that definition of uniformly continious functions applies to my specific f function?

    Best Regards,

    Fred

     
  15. Feb 8, 2006 #14

    benorin

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    What is the definition of a uniformly continuous function?
     
  16. Feb 8, 2006 #15
    Definition:uniformly continuous function

    S is a subset of [tex]\mathbb{R}^n[/tex] and [tex]f: S \rightarrow \mathbb{R}[/tex] be a real valued function. The function f is said to be uniformly continuous, if there for every epsilon > 0, exists a delta > 0 such that,

    |f(x) - f(y) | < epsilon

    for all x,y \in S, which statisfies that ||x - y|| < delta

    Do I then apply this definition to my given f(x) ? and use it to show how f(x) = 0, if x belongs to Y ?

    Best Regards
    Fred

     
  17. Feb 8, 2006 #16

    benorin

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    Yes, if you wish to prove f is a uniformly continuous function do just that: the first bit is done, so just choose delta = epsilon and it is done.

    And do U mean x belongs to S? Indeed it is a simple matter that f(x)=0 for every x in S, since (for fixed x) the shortest distance between x and y when y is allowed to be x is 0.
     
  18. Feb 8, 2006 #17
    regarding uniform continuity

    then according to the definition of uniformly continious functions then
    [tex]|x - y| = |x| - |y|[/tex]

    From here I'm a bit unsure

    Do I do the following y = x + [tex]\delta[/tex] then [tex]x \geq 0 [/tex]

    [tex]|x - (x+ \delta)| = \delta[/tex]

    then [tex]|x - (x+ \delta)| < \epsilon[/tex]

    futermore since [tex]\delta > 0[/tex] according to the definition:

    [tex]\delta < \epsilon[/tex]

    therefore f is uniformly continious.

    Am I on the right path here?

    Best Regards
    Fred
     
    Last edited: Feb 8, 2006
  19. Feb 8, 2006 #18

    benorin

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    Recall that we just proved that

    [tex] |f(x) - f(y) | \leq \| x-y\| [/tex] for all [tex]x,y\in\mathbb{R}^n[/tex]

    we now wish to show that

    [tex]\mbox{ For every } \epsilon >0, \mbox{ there exists a } \delta >0\mbox{ such that }\| x-y\| < \delta \Rightarrow |f(x)-f(y)| <\epsilon [/tex]

    So fix [tex]\epsilon >0[/tex]. Choose [tex] \delta = \epsilon [/tex] so that

    [tex]\| x-y\| < \delta \Rightarrow |f(x)-f(y)| \leq \| x-y\| < \delta =\epsilon [/tex]

    where we have used that bit of which I wrote "Recall that we just proved ..." to get the [tex] |f(x)-f(y)| \leq \| x-y\| [/tex] part of the above line.

    --Ben
     
    Last edited: Feb 8, 2006
  20. Feb 8, 2006 #19
    Thank You again,

    If have two small final followup questions:

    (a) If S is closed set, and if x \notin S, then f(x) > 0

    (b) if Y is closed, then

    X = {x \in R^n | f(x) = 0}

    I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero, but does that explain f(x) > 0??

    Any hints on how to solve (b) ?

    Best Regards

    Fred

     
  21. Feb 8, 2006 #20

    benorin

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    "I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero, but does that explain f(x) > 0??"

    How do you know this is so? If had, say, S=(0,1) (the segment of the real line strictly between 0 and 1) what than is f(0) ?
     
  22. Feb 8, 2006 #21
    If S = (x=1, y = 0) then f(x) > 0 ??

    I'm a bit lost here, any hints?

    Sincerely
    Fred
     
  23. Feb 9, 2006 #22

    benorin

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    This is a trick question:

    I was trying to ensure you understood why "I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero..."

    Understand that if y is the particular point in S whose distance to x is a minimum, then f(x)-||x-y||, that is to say the value of f(x) is that minimum distance between x and y.

    A geometric example will help: We're in [tex]\mathbb{R}^2[/tex]. Let S be the unit square [0,1]x[0,1]. Now fix a point x in [tex]\mathbb{R}^2[/tex], I'll pick x=(2,1/2). Now f[(2,1/2)] can be determined as follows:

    first, find the point in S (the unit square) which is closest to (2,1/2) (which is x in our example), and denote this "closest point" by y. So what is y? Well clearly y is on the right-hand edge of the unit square S since were it not, we could easily find a point in S that is closer to (2,1/2). Every point on the right-hand edge of S is of the form (1,t), where [tex]0\leq t\leq 1[/tex]. The distance [tex]d[/tex] between (1,t) and (2,1/2) is given by the usual distance formula

    [tex]d=\sqrt{(2-1)^2+(1/2 -t)^2}=\sqrt{1+(1/2 -t)^2}[/tex]

    and we seek the value of t with [tex]0\leq t\leq 1[/tex] that minimizes [tex]d[/tex]. We reason that the minimum occurs for t=1/2 and hence determine that y=(1,1/2).

    Second, we recall that f(x) is the distance between x and y, i.e. f(x)=||x-y||, and, in this example, the distance is given by the usual formula, so we have

    [tex]f[(2,1/2)] = ||(2,1/2)-(1,1/2)||=\sqrt{(1-2)^2+(1/2 -1/2)^2}=1[/tex].

    Which makes sense, if you think about the diagram (you should draw one). End Geometric Example.

    Notice that in the above example S was a closed set. In the so-called trick question I posed at the top of this post, S was (0,1) on the real line (which is not a closed set) and I had asked what then is f(0)? We would first determine the point y in (0,1) whose distance to 0 is a minimum, at least we would if there were such a point, but 0 is a limit point of (0,1) so there are infinitely many points in (0,1) infinitely close to 0, but there is always one closer... strictly speaking, there is no point y in (0,1) of minimum distance to 0, but there is an infimum (the inf) of the set of such distances, which is 0 (this is the value of the distance, not the point y at which it occurs), and thus f(0)=0. But notice that 0 (the point) is not in S, and yet we have f(0)=0 and not f(0)>0. Why?


    Answer: S is not closed. In the example where S was the unit square in R^2, S was closed. Can you find a point z in R^2 that is not is S such that f(z)=0? Why not?
     
    Last edited: Feb 9, 2006
  24. Feb 16, 2006 #23
    Hi all,

    I been thinking regarding:

    (a) If S is closed set, and if [tex]x \notin S[/tex], then f(x) > 0

    (b) if Y is closed, then

    [tex]X = \{x \in R^n | f(x) = 0 \}[/tex]

    I know that according to my textbook the definition of a closed set is a as follows:

    A subset F of [tex]\mathbb{R}^n[/tex] is closed if its complement:

    \mathbb{R}^n \mathrm{\} [tex] F = \{ x \in R^n | x \notin F \}[/tex] is open.

    Do I then need to show that x is an internal point of [tex]\mathbb{R}^n[/tex]


    Sincerely

    Fred
     
    Last edited: Feb 16, 2006
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