Dear Colleagues,
I have a problem in the attachment and could you please help me in solving it.
Best Regards.
Raed.
Since $\displaystyle \mathbb{R}^n$ is an inner-product space, every linear functional is given by an inner product. In fact, if $\displaystyle \alpha = (\alpha_1,\alpha_2)$ then the functional $\displaystyle f(x) = \alpha_1\xi_1+\alpha_2\xi_2$ is given by $\displaystyle f(x) = \langle x,\alpha\rangle$. Use that fact to show that $\displaystyle \|f\|_{\mathbb{R}^2} = \sqrt{\alpha_1^2+\alpha_2^2}$.
Now that you know $\displaystyle \|f\|$, you have to figure out how to extend f to a linear functional on $\displaystyle \mathbb{R}^3$ without increasing the norm.
So the extension to $\displaystyle \mathbb{R}^3$ must be of the form $\displaystyle \tilde{f}(x) = \alpha_1\xi_1+\alpha_2\xi_2+\alpha_3\xi_3$, and its norm will be $\displaystyle \|\tilde{f}\| = \sqrt{\alpha_1^2+\alpha_2^2+\alpha_3^2}.$ That leaves only one choice for $\displaystyle \alpha_3.$