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  1. #1
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    extension of functional

    Hello! I have got problems with the following exercise: Let f be the continuous linear functional defined on the subspace U=\{(x,y,z): x+2y=0, z=0\} in the Banach space (\mathbb{R}^3,||.||_1) with f((x,y,z)):=x. Find different linear extensions on the whole space with same norm as f!

    I have calculated the norm of f on U, it is  \frac{2}{3}. My first idea was to extend (2,-1,0) to the basis v_1=(2,-1,0,), v_2=(0,1,0), v_3=(0,0,1) and define f(v_2)=f(v_3)=0.

    But does this extension have the same norm? If yes, how can i show this? How do i find another extension? Hoping for your advice,

    Recursion
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  2. #2
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    Quote Originally Posted by Recursion View Post
    Let f be the continuous linear functional defined on the subspace U=\{(x,y,z): x+2y=0, z=0\} in the Banach space (\mathbb{R}^3,||.||_1) with f((x,y,z)):=x. Find different linear extensions on the whole space with same norm as f!

    I have calculated the norm of f on U, it is  \frac{2}{3}.
    This looks as though it is meant to be an exercise in understanding the proof of the Hahn–Banach theorem. So take a look at that proof, and see if you can mimic its construction.

    You start with f defined on the 1-dimensional subspace U, and you extend it one dimension at a time to the whole space. Vectors in U are of the form (–2y,y,0), and f(–2y,y,0) = –2y.

    So as the first step, let's try to extend f to the subspace consisting of all vectors of the form (x,y,0). To do this, we have to assign a value α to f(1,0,0). Then (x,y,0) = (x+2y,0,0) + (–2y,y,0), and so f(x,y,0) = \alpha(x+2y) -2y = x\alpha + y(2\alpha-2).

    We have to ensure that |f(x,y,0)|\leqslant \tfrac23\|(x,y,0)\|_1 = \tfrac23(|x|+|y|) (because we have to keep the norm of the extension down to 2/3). You can convert that into a condition on α (just as in the proof of the H–B theorem), and I think you'll find that α has to have the unique value 2/3. Thus f(x,y,0) = \tfrac23(x-y).

    Now you have to extend f one dimension further, to the whole space. That means that you have to find a value β for f(0,0,1), without increasing the norm of f. I'll leave the details to you, but I think you'll find that β is not uniquely determined. In that way, you will be able to find different linear extensions of f with the same norm 2/3.
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  3. #3
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    Thanks for your answer. I do not quite get how i can use the proof as the value for the extension is not really determined there. Can you please show me how it follows that a is 2/3?
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  4. #4
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    Quote Originally Posted by Recursion View Post
    Thanks for your answer. I do not quite get how i can use the proof as the value for the extension is not really determined there. Can you please show me how it follows that a is 2/3?
    We need to define \alpha := f(1,0,0) in such a way that f(x,y,0)\leqslant\tfrac23\|(x,y,0)\|_1, or in other words \alpha(x+2y)-2y\leqslant\tfrac23(|x|+|y|) for all x and y.

    Put x=1 and y=0 and the equation tells you that \alpha\leqslant\tfrac23. Put x=0 and y=–1, and it says 2-2\alpha\leqslant\tfrac23, or \alpha\geqslant\tfrac23. Therefore \alpha=\tfrac23 is the only possibility.
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  5. #5
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    Oh yeah, thank you very much! I did not see that .
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