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Estimation of the Operator norm

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data

    L : R^n → R is defined L(x1 , . . . , xn ) = sum (xj) from j=1 to n.

    The problem statement asks me to find an estimation for the operation norm of L, where
    on R the norm ll . llp, 1 ≤ p ≤ ∞, is used and on R the absolute value.


    3. The attempt at a solution

    from,

    ll Lv lly ≤ M ll v llx

    I plan to find the smallest ''M'' since that's the operator norm.

    I assume that the linear operator is continuous.

    The problem statement mentions the Hölder inequality - as a hint.

    The truth is that I'm not sure what exactly I have to do in order to show what I plan to do. I have tried to do something but get stucked at the very begining...

    Any help will be very appreciated.
     
  2. jcsd
  3. May 13, 2012 #2
    The fact that the problem asks you to estimate the operator norm is an indication that this plan is unrealistic. You're just meant to find a "good" upper bound for ## M ##.

    Not that it's part of the problem, but you should be able to prove (easily) that ## L ## is continuous and linear.

    This is a big part of the solution. If I give you ## \sum |x_k|=\sum |x_k\cdot 1| ## and tell you to apply Holder to the right hand side, is that enough to get you started?
     
  4. May 13, 2012 #3
    Hello. Thanks for your answer!

    Well, I applied Hölder's inequality. Then I found that the p-norm of L times the p-norm of 1 is bigger than L. Then by the definition of the operator norm, I can say that the right hand of the inequality is indeed the operator norm?
     
  5. May 13, 2012 #4
    Close. It sounds like you're getting your norms mixed up. In this case, we have the usual norm ## |\cdot| ## on ## \mathbf{R} ##. There is a norm ## \|\cdot\|_p ## on ## \mathbf{R}^n ## defined by ## \|\mathbf{x}\|_p=\left(\sum|x_k|^p\right)^{\frac{1}{p}} ##. There is also the (induced by ## p ##) operator norm ## \|\cdot\|_{Op} ## defined on the set of bounded linear operators. For a bounded linear operator ## L:\mathbf{R}^n\rightarrow\mathbf{R} ##, ##\|L\|_{Op}=\min\{M: |L(\mathbf{x})|\leq M\cdot\|\mathbf{x}\|_p ## for all ## \mathbf{x}\} ##. Note all of the different norms going on in this definition. You really gotta pay attention to what you're doing here.

    Also, it looks like you're a bit mixed up on Holder's inequality, which states the following:

    For ## \frac{1}{p}+\frac{1}{q}=1 ##, ## \sum |x_k\cdot y_k|\leq\left(\sum|x_k|^p\right)^{\frac{1}{p}} \cdot \left(\sum|y_k|^q\right)^{\frac{1}{q}} ##


    AVERT YOUR EYES IF YOU WISH TO FINISH ON YOUR OWN


    Here's how the solution should look.

    For a fixed ## p ##, let ## q ## be such that ## \frac{1}{p}+\frac{1}{q}=1 ##.

    Then for all ## \mathbf{x} ##,

    ## |L(\mathbf{x})|=|\sum x_k|\leq_{\triangle}\sum|x_k|= \sum |x_k\cdot1|\leq_H\left(\sum|x_k|^p\right)^{\frac{1}{p}} \cdot \left(\sum|1|^q\right)^{\frac{1}{q}}=\|\mathbf{x} \|_p\cdot n^{\frac{1}{q}}= n^{1-\frac{1}{p}}\cdot\|\mathbf{x}\|_p##, where ##\leq_\triangle## and ##\leq_H## indicate inequality due to triangle and Holder inequalities respectively.

    So ##\|L\|_{Op}\leq n^{1-\frac{1}{p}}##.
     
    Last edited: May 13, 2012
  6. May 14, 2012 #5
    Thanks. I’ll try to be more aware with the meaning of each part of a definition.

    But I’m still not sure what’s going on. The definition says that the L_op is defined as the min M such that the norm of L(x) on R for all x is less or equal than this M times the p-norm on R^n. So the length of L(x) (in R) is less than, eh, the lengh of x in R^n times M? And M is just a scalar, right? if I would like to see it in a graph, for example, how could I represent it?

    Induced by p?
     
  7. May 14, 2012 #6
    Yeah. You really gotta pay attention to the definitions. And recognize that they're "just" definitions; it is what it is for no other reason than because we say so.

    Of course most definitions in mathematics have a purpose. So there really is a reason beyond "because we say so". It's usually more along the lines of "because that's what works".

    Yes. And again, this is just the definition of the operator norm. There are other norms that we could choose to measure the "size" of functions from one space to another. This one just happens to be one that works for us in some cases.

    Yeah ... no ... unless you can visualize the Cartesian product ##\mathcal{L}\times\mathbf{R}^+## of the space ##\mathcal{L}## of linear operators ##L:\mathbf{R}^n\rightarrow\mathbf{R}## with the non-negative reals. In which case you should maybe consider a second career in art.

    It's OK to try to visualize some mathematics. But in almost all but the most basic cases, it's not really possible. This is one of those cases where you just gotta say to yourself, "Hey, I know intuitively what 'size' is supposed to mean. This norm is just one way of measuring the 'size' of some object that I can't see."

    Maybe not the best choice of words on my part. I just meant that it's the operator norm on bounded linear functions ##L:\mathbf{R}^n\rightarrow\mathbf{R}## where the norm on ##\mathbf{R}^n## is the ##p##-norm. Operator norms are dependent on the norms chosen for the two vector spaces which are the domain and range of the operators.
     
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