# Extending a functional with same norm

I have: $X = <\mathbb R^2, ||(x,y)|| = \sqrt{4x^2+y^2}>, L = \{(2x,3x), x \in\mathbb R\},$

$\phi_0 \in L^* : \phi_0 (2x,3x) = -2x$

I need to extend $\phi_0$ on $X$ without changing norm (norm should stay the same). How can I do it?

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i think by "expand" you mean "extend".......... –  Fermat May 13 at 7:41
@Hamid Oh, thanks –  michaeluskov May 13 at 7:45
I'm not clear what you're doing on line 2: the domain of $\phi_0$ is elements of the form $(2x,3x)$. How is $(0,2x)$ of this form? Is this a typo? –  Matt N. May 13 at 10:21
@MattN. oh, this is mistake. I edited post. –  michaeluskov May 13 at 14:42

First way of solving the question:

To get the existence of an extension you may apply the Hahn-Banach extension theorem. To do so you will need a sublinear map on $X$ that dominates $\phi_0$ on $L$.

Since as it's written $\phi_0$ is not defined on $L$ I'll assume you meant $$\phi_0 (2x,3x) = -2x$$

To find a sublinear function $p:X \to \mathbb R$ you can choose for example the norm. Then

$$\|(x,y)\| = \sqrt{4x^2 + y^2} \ge \sqrt{4x^2} = 2 |x| \ge -2x = \phi_0(2x,3x)$$

Since $\phi_0$ is dominated by $\|\cdot \|$ on $L$ it follows from Hahn-Banach that there exists an extension with same norm.

Second (constructive) way of solving the question:

You could define $\phi : X \to \mathbb R, (x,y) \mapsto -x$. This map extends $\phi$ and is continuous. Now all you need to do is show that it has the same norm as $\phi_0$.

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So, my extension is $\phi_1(2x,3x) = \sqrt{4x^2+y^2}$? –  michaeluskov May 13 at 14:57
@user23791 No, that's your sublinear map bounding $\phi_0$. Hahn-Banach doesn't construct an extension, it gives you its existence. –  Matt N. May 13 at 15:24
But how to construct an extension? –  michaeluskov May 13 at 15:27
@user23791 You don't need to construct it explicitly. –  Matt N. May 13 at 15:29
@user23791 But you can. I'll add a second way of solving the question. –  Matt N. May 13 at 15:31

Note first that $X$ is an inner-product space. The norm is given by the inner product $$\langle (x,y),(w,z)\rangle = 4xw+yz.$$

Let us calculate the norm of $\phi_0$. We have $$\|\phi_0\|=\sup\left\{\frac{|\phi_0(2x,3x)|}{\|(2x,3x)\|}:\ x\ne0 \right\} =\sup\left\{\frac{|-2x|}{\sqrt{16x^2+9x^2}}:\ x\ne0 \right\} =\sup\left\{\frac{2|x|}{5\,\sqrt{x^2}}:\ x\ne0\right\}=\frac25.$$ If we try to extend $\phi_0$ as $\phi(x,y)=-x$, then the norm is not preserved: $$\|\phi\|=\sup\left\{\frac{|\phi(x,y)|}{\|(x,y)\|}:\ (x,y)\ne(0,0) \right\} =\sup\left\{\frac{|-2x|}{\sqrt{4x^2+y^2}}:\ (x,y)\ne(0,0) \right\} =\sup\left\{\frac{2|x|}{\sqrt{4x^2+y^2}}:\ (x,y)\ne(0,0) \right\}=1$$ (the sup is attained when $y=0$).

The way to extend $\phi_0$ without enlarging the norm is to define it as zero on the orthogonal complement of $L^*$. As the dimension of $X$ is $2$ and the dimension of $L^*$ is $1$, the dimension of the orthogonal is also $1$. If $(z,w)\perp(2x,3x)$, we have $$0=\langle(z,w),(2x,3x)\rangle=8xz+3xw=(8z+3w)x.$$ As this should be zero for all $x$, we get $8z+3w=0$, or $w=-8z/3$. So a basis for $(L^*)^\perp$ is given by $\{(3,-8)\}$. Then we can define $$\phi(\alpha (2,3)+\beta(3,-8))=-2\alpha.$$ If we want to express $\phi$ in terms of a single pair $(x,y)$, we need to write $$(x,y)=\frac{8x+3y}{25}\,(2,3)+\frac{3x-2y}{25}\,(3,-8),$$ so $$\phi(x,y)=-2\,\frac{8x+3y}{25}=\frac{-16x-6y}{25}.$$ A little multivariable calculus can show that $\|\phi\|=2/5$, as desired.

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