For a subspace $E \subset V$ and a linear functional $f \in E^{*}$ with $||f||=1$ there is a extension $g \in span\{v, E\}^{*}$ where $v \not \in E$

Question

Let $V$ be a normic space and $E \subset V$ a subspace. Let $v \in V \setminus E$. Let $f \in E^{*}$ be a bounded functional such that $||f||=1$.

I declare $W = span\{v,E\}$ and want to show that there is $g \in W^{*}$ such that $g|_E = f$ and $||g||=1$.

I started by taking $x \in W$, then there is a scalar $\lambda$ and $e \in E$ such that $x = \lambda v + e$ and such $\lambda$ and $e$ are unique.

Then I declared the functional $g(x) = g(\lambda v + e) = \lambda + f(e)$.

If $x \in E$ then $\lambda = 0$ so we get $g(x) = g(e) = f(e)$ concluding that $g|_E = f$. The only thing left is to show that $||g||=1$.

I managed to get $||g||\ge1$ since $$sup_{||\lambda v + e||=1} |g(\lambda v + e)| = sup_{||\lambda v + e||=1} |\lambda + f(e)| \ge sup_{||e||=1} |f(e)| = ||f|| = 1$$

However, I couldn't get the inequality $||g|| \le 1$.

Help would be appreciated.