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Homework Help: Smallest Norm in a Hilbert Space

  1. Apr 25, 2005 #1
    I have this problem which I want to do before I go back to uni. The context was not covered in class before the break, but I want to get my head around the problem before we resume classes. So any help on this is greatly appreciated.

    Question

    Suppose [itex]C[/itex] is a nonempty closed convex set in a Hilbert Space [itex]H[/itex].

    (i) Prove that there exists a unique point [itex]c_0 \in C[/itex] of smallest norm, and that we then have [itex]\Re \langle c_0 \, | \, c-c_0\rangle \geq 0[/itex] for all [itex]c \in C[/itex].

    (ii) For any point [itex]x_0 \in H[/itex] show that there is a unique closest point [itex]c_0[/itex] of [itex]C[/itex] to [itex]x_0[/itex], and that it satisfies the
    variational inequality [itex]\Re \langle x_0 - c_0 \,|\, c -c_0\rangle \leq 0[/itex] for all [itex]c \in C[/itex]
     
    Last edited: Apr 25, 2005
  2. jcsd
  3. Apr 25, 2005 #2
    Solution (i)

    Since we are in a Hilbert space I figured that using the distance metric and perhaps the parallelogram law will be used.

    Let [itex]d[/itex] be the smallest distance from [itex]C[/itex] to [itex]c[/itex]. This can also be written as

    [tex]d = \inf d(x,y)[/tex]

    The first thing I thought was, how can I make such a bold statement? Does [itex]d[/itex] even exist? But I assured myself that there is such a [itex]d[/itex] because we made the condition that [itex]C[/itex] was a closed convex set, therefore we can make [itex]d[/itex] as small as we like. If [itex]C[/itex] wasn't closed convex then there would be trouble at the boundaries.

    Anyway, the next step involves setting up a sequence of elements of [itex]C[/itex], like the sequence of [itex]y_n \in C[/itex]. This enables us to introduce the parallelogram law.

    [tex]\|c - y_n\| = d(x,y_n) \rightarrow d[/tex].

    So the distance from [itex]c[/itex] to every element [itex]y_n[/itex] in the sequence approaches the smallest distance [itex]d[/itex] the further along the sequence we go. But this can only happen if [itex](y_n)[/itex] is a Cauchy sequence. We know that Cauchy sequences behave like this in Hilbert Spaces, and the easiest way I know of proving [itex](y_n)[/itex] is a Cauchy sequence is to use the Parallelogram Law and prove that it approaches 0 as [itex]n \rightarrow 0[/itex]. So we have

    [tex]\|y_n - y_m\|^2 &=& 2\|y_n - c\|^2 + 2\|y_m - c\|^2 - \|(y_n+y_m)-2c\|^2 [/tex]

    [tex]
    \leq 2\|y_n - c\|^2 + 2\|y_m - c\|^2 - 4d^2 [/tex]

    [tex]
    \rightarrow 0[/tex]

    Remember that [itex]d[/itex] is our minimum distance. So we know [itex](y_n)[/itex] is a Cauchy sequence. And we also know that [itex]C[/itex] is closed. Hence [itex]C[/itex] is a complete space and so [itex]y_n \rightarrow c_0 \in C[/itex]. That is, the Cauchy sequence approaches the unique point [itex]c_0[/itex] as [itex]n \rightarrow \infty[/itex].

    It follows, by the continuity of the norm, that

    [tex]\|c-c_0\| = d[/tex]

    Which says that there exists a distance [itex]d[/itex] from any arbitrary point [itex]c \in C[/itex] to the point [itex]c_0 \in C[/itex] that is minimum, that is there exists a smallest norm.
     
    Last edited: Apr 25, 2005
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