I am looking for a proof of the following theorem. The statement is not necessarily accurate, so please don't quote from me.
Let $X$ be a Banach space. $\dim X=\infty$. $E\subseteq X$ is a finite dimensional subspace. For all $\epsilon>0$ there exist $x\in X$ with $\|x\|=1$, such that for all $y\in E,\lambda\in \mathbb F$(that is $\mathbb R$ or $\mathbb C$), $\|y\|\leq(1+\epsilon)\|y+\lambda x\|$.
This theorem does not have a name. (Or I have forgotten the name.) I struggle to find it. Can anyone offer me a reference or a proof?
I have figured out a proof myself. Since $E$ is finite-dimensional, the close unit sphere $S=\{x\in E: \|x\|=1\}$ is compact, and thus totally bounded. Therefore, we have, for any $\epsilon>0$, a finite set of points $\{y_i\}_{i=1}^n$ such that for all $y\in S$, there exists $1\leq i\leq n$ such that $\|y-y_i\|<\epsilon/2$.
Now, consider the linear functional $f_i(\lambda y_i)=\lambda$ defined on $\text{span }y_i$. $\|f_i\|=1$ and $f_i$ is bounded by the norm $\|\cdot\|$, so by Hahn-Banach Theorem, we can extend $f_i$ to be defined on $X$, i.e., $f_i\in X^*,\|f_i\|=1,f_i(y_i)=1$.
The null space or kernel of $f_i$, $\ker f_i$, is a maximal proper subspace of $X$. Therefore, $\dim\ker f_i=\infty$. We now prove that $\dim( \bigcap_{i=1}^m \ker f_i)=\infty$ for all $1\leq m\leq n$. We prove by induction on $m$. This is obviously true for $m=1$. Now suppose by induction assumption that $\dim( \bigcap_{i=1}^{m-1} \ker f_i)=\infty$, and let $K_{m-1}=\bigcap_{i=1}^{m-1} \ker f_i$. If $K_{m-1}\subseteq \ker f_m$, then obviously $\dim K_m=\infty$. Otherwise, $K_m=K_{m-1}\cap \ker f_m$ is a maximal proper subspace of $K_{m-1}$ or $K_m=K_{m-1}$. In both case, it is easy to show that $K_m$ is infinite-dimensional.
Therefore, space $K_n=\bigcap_{i=1}^{n} \ker f_i$ is infinite-dimensional and therefore non-zero. So, there exist $x\in K_n$ with $\|x\|=1$. So, we have $f_i(x)=0$ for all $i$.
Now, for $y\in E$ and $\|y\|=1$ we have $$ \|y+\lambda_x\|\geq \|y_i+\lambda x\|-\frac{\epsilon}{2}\geq f_i(y_i+\lambda x)-\frac{\epsilon}{2}=1-\epsilon/2\geq \frac{\|y\|}{1+\epsilon} $$
If $\|y\|\neq 1$, just substitute $y'=y/\|y\|$. If $y=0$, then the theorem is obviously true. So the proof is complete.