extension of functional

**Recursion** · January 5th 2009, 07:52 AM

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

**Opalg** · January 5th 2009, 11:35 AM

Originally Posted by Recursion

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.

**Recursion** · January 5th 2009, 01:52 PM

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?

**Opalg** · January 5th 2009, 11:56 PM

Originally Posted by Recursion

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.

**Recursion** · January 13th 2009, 12:17 PM

Oh yeah, thank you very much! I did not see that

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