Suppose $E$ is a finite-dimensional subspace of a Banach space $X$, and $x_0\in X$ is a vector with $x_0\notin E$. Suppose $f\in X^*$ is a continuous linear functional and that $a>0$ is a positive real number.
Conjecture 1. There exists a continuous linear functional $g\in X^*$ with $g(e)=f(e)$ for all $e\in E$ and $0<g(x_0)<a$, and such that $\|g\|\leq\|f\|$.
If this is impossible, the following weaker conjecture will also do for my purposes.
Conjecture 2. For every $\epsilon>0$ there exists a continuous linear functional $g\in X^*$ with $g\equiv f$ on $E$ and $0<g(x_0)<a$, and such that $\|g\|\leq\|f\|+\epsilon$.
Via Hahn-Banach we can assume that $X$ has dimension exactly one greater than $E$, so that $g$ is defined by setting $g(e)=f(e)$ for all $e\in E$ and finding a value $0<g(x_0)<a$ which guarantees $\|g\|\leq\|f\|$. If true, this should be a simple functional analysis exercise. However, I am not seeing right now how to do it. Maybe this is just some form of the separation theorem.
Thanks!
Neither conjecture is true. Conjecture 1 fails trivially if $f=0$. For a less trivial counterexample that also violates Conjecture 2, let $X=\mathbb{R}^2$ with the sup norm, let $E=\{(x,0):x\in\mathbb{R}\}\subset X$, and let $f(x,y)=x$. Then $\|f\|=1$. Now take $x_0=(-2,1)$. If $g\in X^*$ agrees with $f$ on $E$ and is such that $g(x_0)>0$, then $$g(0,1)=g(x_0)+g(2,0)>g(2,0)=f(2,0)=2.$$ Thus for any such $g$, $\|g\|>2$.