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Square-integrable functions as a vector space

  1. May 21, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Show that the set of all square-integrable functions is a vector space. Is the set of all normalised functions a vector space?

    (b) Show that the integral ##\int^{a}_{b} f(x)^{*} g(x) dx## satisfies the conditions for an inner product.

    2. Relevant equations

    The main problem is to show that the sum of two square-integrable functions is itself square-integrable.

    Use Schwarz inequality: ##| \int^{a}_{b} f(x)^{*} g(x) dx| \leq \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx}##

    3. The attempt at a solution

    (a) Let ##f(x)## and ##g(x)## be square-integrable functions. Then,

    ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx##

    ##= \int^{a}_{b} [\ | \alpha f(x) |^2 + | \beta f(x) |^2 + \alpha^{*} f(x)^{*} \beta g(x) + \alpha f(x) \beta^{*} g(x)^{*} ] dx ##

    ##= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + \alpha \beta^{*} \int^{a}_{b} f(x) g(x)^{*} dx ##

    The first two terms are already square-integrable. So, all I have to do now is to show that the last two terms are square-integrable.

    Can you suggest how I can use the Schwarz inequality over the last two terms?

    P.S. : The last two terms are complex conjugates of each other, and the sum of two complex conjugates is a real number. Does it somehow prove that the last two terms form a square-integrable combination?
     
    Last edited: May 21, 2015
  2. jcsd
  3. May 21, 2015 #2

    Orodruin

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    The last two terms together are real. For any complex number ##z##, ##{\rm Re}(z) \leq |z|##. Then apply the Schwarz inequality.
     
  4. May 21, 2015 #3
    Alright, here's my attempt.

    Let ##f(x)## and ##g(x)## be square-integrable functions. Then,

    ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx##

    ##= \int^{a}_{b} [\ | \alpha f(x) |^2 + | \beta f(x) |^2 + \alpha^{*} f(x)^{*} \beta g(x) + \alpha f(x) \beta^{*} g(x)^{*} ] dx ##

    ##= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + \alpha \beta^{*} \int^{a}_{b} f(x) g(x)^{*} dx ##

    ##= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + ( \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx )^* ##

    ##= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ {\rm Re} ( \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx ) ##

    ##\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx | ##

    ##\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta | \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx} ##

    This is a combination of square-integrable functions, so that ##\alpha f(x) + \beta g(x)## is a square-integrable function.


    Let ##\int^{a}_{b} |f(x)|^2 dx = 1## and ##\int^{a}_{b} |g(x)|^2 dx = 1##. Therefore,

    ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx##

    ##\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta | \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx} ##

    ##\leq |\alpha|^2 + |\beta|^2 + 2\ | \alpha^{*} \beta |##

    ##\leq |\alpha|^2 + \alpha^{*} \beta + \alpha \beta^{*} + |\beta|^2##

    ##\leq | \alpha + \beta |^2##.

    How can this help to show that the set of all normalised functions is a vector space?
     
  5. May 21, 2015 #4

    Orodruin

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    This shows you that any (finite) linear combination of functions in the space is also in the space. You should check the other axioms of a vector space as well, but it is rather trivial.
     
  6. May 21, 2015 #5
    How exactly can this show any (finite) linear combination of functions in this space is also in this space?

    For one thing, I've only shown that, given ##\int^{a}_{b} |f(x)|^2 dx = 1## and ##\int^{a}_{b} |g(x)|^2 dx = 1##,

    ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \leq | \alpha + \beta |^2##, not that ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx = 1##.

    That does not quite prove that the set of all normalised functions form a vector space, does it (skipping the other axioms of a vector space, which are rather trivial to prove for this case)?
     
  7. May 21, 2015 #6

    Orodruin

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    Ah right, I missed the "normalised". It is trivial to shot that it is not a vector space.
     
  8. May 21, 2015 #7
    Haven't I already shown that the set of all normalised functions is ##\textit{not}## a vector space when I proved the following:

    if ##\int_{a}^{b} |f(x)|^{2} dx = 1## and ##\int_{a}^{b} |g(x)|^{2} dx = 1##, then ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \leq | \alpha + \beta |^2##??
     
  9. May 21, 2015 #8

    Orodruin

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    Almost. You need to insert some values. But that is not the easiest way to prove it. The space has no zero element.
     
  10. May 21, 2015 #9
    Right! I can see that the function ##f(x) = 0## is not normalisable.

    Therefore, lacking a zero element, the set of all normalised functions do not form a vector space.

    I wonder how to complete the proof using my original approach. Is this the correct way?

    Let ##\alpha = 0.1## and ##\beta = 0.1##, then ##|\alpha + \beta|^2## = 0.01.

    So, ##\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \neq 1##.
     
    Last edited: May 21, 2015
  11. May 21, 2015 #10

    Orodruin

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    Yes, that is fine apart from that the ##|\alpha + \beta|^2## = 0.01 should be an inequality.
     
  12. May 21, 2015 #11

    Ray Vickson

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    Even easier: if ##f## is normalized to 1 and ##c## is a scalar with ##|c| \neq1##, then ##c f## is not normalized to 1.
     
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