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?
 A: 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.
