I want to find a Banach space $E$ and a compact operator $K:[0,1]\times E \rightarrow E$ (that is, $K$ maps every bounded sequence onto a sequence that converges up to a subsequence) satisfying the following conditions:


1. $K(0,\cdot) = 0$
2. There is a $r>0$ and a sequence $(\lambda\_n,u\_n)\in [0,1]\times \overline{B}\_r(0)$ such that $\lambda\_n\rightarrow 0$ but, for each $N\in \mathbb{N}$, it is possible to find $n>N$ such that $K(\lambda\_n,u\_n)\not\in B\_r(0)$.


My attempt: let $E=c\_0$ endowed with the maximum norm, where $c\_0$ is the Banach space of the sequences that converges to zero. Consider the operator $K:[0,1]\times c\_0\rightarrow c\_0$ defined by


$$K(\lambda,u)=2(\lambda u\_1,\lambda^{1/2} u\_2^2,\ldots,\lambda^{1/n} u\_n^n).$$
If we take $r=1$, then we have


$$ K(1/n,e\_n) = \frac{2}{n^{1/n}}\rightarrow 2>1=r.$$


The problem with my attempt is that apparently the operator $K$ is not compact.



Example of a compact operator that is not uniformly continuous
Your idea almost works.


Let $E = \ell\_1$ be the space of absolutely summable sequences.
Let
$$ K(\lambda, (u\_k)\_{k\in\mathbb{N}}) = (2\sum \lambda^{1/k} |u\_k|, 0, 0, \ldots) $$


Given any bounded set in $E$, its image is (essentially) a bounded set in $\mathbb{R}$, and so is pre-compact.


$K(0,\cdot)$ is obviously $0$.


Letting $e\_n$ be the element of $\ell\_1$ that has a $1$ on the $n$th spot and zero otherwise, you get


$$ K(1/n, e\_n) = (2 (1/n)^{1/n}, 0, 0, 0, \ldots) $$


which is eventually outside $B\_1(0)$.


