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Math Help - integrals squared

  1. #1
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    integrals squared

    I'm working through a proof for part of my university project and I'm a little stuck understanding part of it:
    if:
    \int_0^1 (f(x))^2 dx =\left(\int_0^1 f(x)dx \right)^2<br />
    then f(x) = constant
    why does this prove f(x) is constant?
    Thankyou
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  2. #2
    GJA
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    Re: integrals squared

    Hi hannah,

    We can get what you want by applying the Cauchy-Schwarz inequality (see Cauchy). The wikipedia article gives the statement of the theorem in terms of arbitrary inner-product spaces; if you know what these are and how they apply to what you've asked about, then read no further. If not I have provided an explanation below about how it helps us establish the result.

    Real-valued continuous functions defined on [0,1] form a vector space over the real numbers (meaning if we add two real-valued continuous functions defined on [0,1] we again get a continuous real-valued function on [0,1], and if we multiply a real-valued continuous function defined on [0,1] by a real scalar we get another real-valued continuous function defined on [0,1]); this vector space is typically denoted by C([0,1], \mathbb{R}). When we have a vector space over a field it is often desirable to attach what is known as an inner-product to the vector space. The inner-product allows us to define geometric quantities on our space like lengths of vectors, angles between vectors, etc. The "dot product" you learned about in multivariable Calculus is an inner-product on \mathbb{R}^{3}, for example. The (standard) inner product for C([0,1], \mathbb{R}) is given by

    <f,g>=\int_{0}^{1}f(x)g(x)dx,

    where f(x),g(x)\in C([0,1],\mathbb{R}).

    Now, as to how all of this pertains to your question, if we take g(x)=1 to be the constant function 1, the Cauchy Schwarz inequality says that

    \left(\int_{0}^{1}f(x)\cdot 1 dx \right)^{2}\leq \int_{0}^{1}f(x)^{2}dx\cdot\int_{0}^{1}1^{2}dx= \int_{0}^{1} f(x)^{2}dx,

    where equality holds above if and only if f(x) and g(x)=1 are linearly dependent. According to the assumption you've provided we do have equality above. Hence, f(x) and g(x)=1 are linearly dependent; i.e. f(x)=c\cdot g(x)=c\cdot 1=c. We see that f(x)=c, so f is constant.

    Does this answer your question? Let me know if anything is unclear. Good luck!
    Last edited by GJA; December 17th 2012 at 12:49 PM.
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  3. #3
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    Re: integrals squared

    Thankyou that is massively helpful, it was the linear dependence bit I was missing, so all makes a lot more sense now!
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