Perturbation of linear Operator Let $X$ be a Banachspace. And $L: X \to X$ a linear bounded Operator. Given $ \varepsilon > 0$  i want to find $\lambda>0 $  such that $\| \lambda I - L \| \le \varepsilon $. Using the Standard Norm for linear Operators.  Is this possible? 
 A: Suppose that to each $ \epsilon >0$ there is $\lambda ( \epsilon)>0$ such that
$||\lambda ( \epsilon)I-L|| \le \epsilon$. Then, for each $n \in  \mathbb N$ there is $\lambda_n >0$ such that
(*) $||\lambda_n I-L|| \le 1/n$.
From $| \lambda_n|=|| \lambda_nI-L+L|| \le || \lambda_nI-L||+ ||L|| \le 1/n+||L|| \le 1+||L||$ we see that $( \lambda_n)$ is bounded. Therefore we can  , without loss of generality, assume that $( \lambda_n)$ converges. Let $\lambda_0$ the limit of this sequence.
Now, from (*) with $ n \to \infty$, we get:
$$L= \lambda_0I.$$
Consequence: an operator $L$ has your above property $ \iff L$ is of the form $\lambda_0I.$
A: The following generalisation of my answer above might be of some interest. The proof should be clear.
Let $E$ be a normed space over $\mathbb K= \mathbb R$ or $\mathbb K= \mathbb C$. For $x,y \in E$ the follwing assertions are equivalent:
$(1)$ for each $ \epsilon >0$ there is $\lambda \in \mathbb K$ with $|| \lambda x-y|| \le \epsilon$;
$(2)$ $x$and $y$ are linearly dependent.
