Give an example to show that extension of the functional of Hahn-Banach is not unique. I tried it that i think of a finite dimensional inner product space.and take it orthonormal basis{v2,v2,...vn} and pick one v1 vector from that .and define f on that vector 1 and so we extend it on the space gendrated by that vector so on this space the norm of the functional is 1.now how i extend it i can't understand..
 A: Consider the space $C[0,1]$ of continuous functions on the unit interval endowed with the supremum norm. Let $E$ be the subspace of constant functions and define a functional $\phi$ by $\phi(f)=f(1)$. Note that $|\phi(f)|=|f(1)|=\|f\|_\infty$ since $E$ consists of the constant functions, so the functional is bounded on $E$ and it has norm one (it is isometric).
Note that there are many functionals on $C[0,1]$ that extend $\phi$. Indeed, for any $t\in[0,1]$, define the evaluation functional $\phi_t:C[0,1]\to\mathbb{C}$ given by $\phi_t(f)=f(t)$. Note that, if $f\in E$ (i.e. $f$ is constant) then $\phi_t(f)=f(t)=f(1)=\phi(f)$, so $\phi_t$ extends $\phi$. Moreover, it is very to see that $\|\phi_t\|=\|\phi\|=1$ (we have $|\phi_t(f)|=|f(t)|\leq\|f\|_\infty$ so $\|\phi_t\|\leq1$ but also $\|\phi_t\|\geq|\phi_t(1)|=1$) , so we have a bunch of isometric extensions of $\phi$ to the entire space $C[0,1]$. This shows that, in general, the norm-preserving extension of a functional described in the Hahn-Banach theorem,, is not unique.
edit
I said something stupid about Hilbert spaces, sorry about that! Of course we can find examples from the Hilbert realm:
Assume that $H$ is a Hilbert space and $K\subset H$ is a linear subspace. We consider $\bar{K}$, the closure, which is a closed subspace, so it is a Hilbert space.
Let $\phi:K\to\mathbb{C}$ be a bounded functional. Since $K$ is dense in $\bar{K}$, $\phi$ extends uniquely to a bounded functional $\bar{\phi}:\bar{K}\to\mathbb{C}$. By the Riesz representation theorem, $\bar{\phi}$ has to be an inner product functional, i.e. there exists a unique $k\in\bar{K}$ so that $\bar\phi(\cdot)=\langle\cdot,k\rangle$. Moreover, $\|\phi\|=\|\bar{\phi}\|=\|k\|$.
Now assume that $\psi:H\to\mathbb{C}$ is a bounded linear functional that extends $\phi$. Then, by the Riesz representation theorem we must have $\psi(\cdot)=\langle\cdot,z\rangle$ for some unique $z\in H$. Therefore we must have that $\langle x,k\rangle=\langle x,z\rangle$ for all $x\in K$. Equivalently, we must have that $\langle x,k-z\rangle=0$ for all $x\in K$.
We know that, if $K$ is not dense in $H$, then $K^\bot$ is a non-zero subspace. Therefore, by picking any $y\in K^\bot$ and setting $z=k+y$ the functional $\phi_z$ defined by $\phi_z(x)=\langle x,z\rangle$ defines an extension of $\phi$ on $H$. Note that $\|\phi_z\|=\|z\|=\sqrt{\|k\|^2+\|y\|^2}$, so the extension is isometric ONLY when $y=0$, i.e. we have a unique isometric extension of $\phi$.
However, if we do not care about having isometric extensions and only care for extensions to bounded functionals, then the above process yields an abundance of such extensions.
A more specific example: take $H=\mathbb{C}^3$ and $K=\{(z,0,0):z\in\mathbb{C}\}$ Then $K$ is a closed subspace and $K^\bot=\{(0,z,w):z,w\in\mathbb{C}\}$. Take the functional $\phi:K\to\mathbb{C}$ given by
$$\phi(z,0,0)=\langle (z,0,0),(1,0,0)\rangle=z$$
Choose any $y,w\in\mathbb{C}$. Then define $\phi_{y,w}:H\to\mathbb{C}$ by
$$\phi_{y,w}(z_1,z_2,z_3)=\langle(z_1,z_2,z_3),(1,y,w)\rangle=z_1+z_2\bar{y}+z_3\bar{w} $$
This is a bounded functional that extends $\phi$.
