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Thread: Norms

  1. #1
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    Norms

    Let $\displaystyle ||-||_{1}$ and $\displaystyle ||-||_{2}$ be the $\displaystyle L^1$-norm and the $\displaystyle L^2$-norm on $\displaystyle C[a,b]$ the space of continuous real valued functions on the closed interval $\displaystyle [a,b]$: explicitly

    $\displaystyle ||f||_{1}=\int_{a}^{b} |f|$, $\displaystyle ||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$.

    (a) Prove that $\displaystyle ||f||_{1} \leq ||f||_{2} $ for all $\displaystyle f \in C[a,b]$ (1). Hint: Remember the Cauchy-Schwartz inequality for integrals.

    (b) For $\displaystyle n \geq 1$, define $\displaystyle f_{n}:[0,1] \to \mathbb{R} $ by

    $\displaystyle f_{n}(x)= \begin{matrix} n & if 0 \leq x \leq \frac{1}{n} \\ & \\ 2n-n^{2}x & if \frac{1}{n} \leq x \leq \frac{2}{n} \\ & \\ 0 & if \frac{2}{n} \leq x \leq 1\end{matrix} $

    Caculate $\displaystyle ||f_{n}||_{1}$ and $\displaystyle ||f_{n}||_{2}$.

    So the thing is that I've been ill for over a week now and couldn't go to lectures resulting in me not having lecture notes and no notion of norms.

    I have found hat Cauchy-Schwartz states that $\displaystyle \int_{a}^{b} f(x)^{2} dx \int_{a}^{b} g(x)^{2} dx \geq (\int_{a}^{b}f(x) g(x) dx)^{2}$ for two real integrable functions in an interval [a,b]. I don't really know how to use this to prove (1).

    As for (b) I am confused by the sequence of functions which I don't really know how to interpret (what is it geometrically?) or plug into $\displaystyle ||f||_{1}=\int_{a}^{b} |f|$, $\displaystyle ||f||_{2}=\sqrt{\int_{a}^{b} |f|^2}$.

    Now, if someone could just start me off that would be great!
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  2. #2
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    Krizalid's Avatar
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    for the first one.

    if $\displaystyle f\equiv0,$ it's a trivial case, so suppose $\displaystyle b>a,$ then $\displaystyle \int_{a}^{b}{\left| f \right|}=\int_{a}^{b}{\left| f \right|\cdot 1}\le \sqrt{\left( \int_{a}^{b}{\left| f \right|^{2}} \right)\left( \int_{a}^{b}{1^2} \right)},$ you can finish that now.
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  3. #3
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    When I do that I end up with $\displaystyle \int_{a}^{b} |f| dx \leq \sqrt{b-a} \sqrt{\int_{a}^{b} |f|^2 dx}$ but this does not necessarily imply $\displaystyle \int_{a}^{b} |f| dx \leq \sqrt{\int_{a}^{b} |f|^2 dx}$, does it?
    I'm really tired so I might be missing something really obvious.
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