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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.

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    $\begingroup$ The second condition also should mention $K$. As it is written now, it does not make sense. $\endgroup$
    – daw
    Sep 19, 2022 at 6:08
  • $\begingroup$ Thank you! It is correct now. $\endgroup$ Sep 19, 2022 at 13:57
  • $\begingroup$ Do the $K(\lambda,\cdot)$ have any connection for different $\lambda$? If not then take any compact operators with operator norm $>1$ for $\lambda>0$. For a sequence $\lambda_n$ take a unit vector $u_n$ that almost achieves this norm. Then $K(\lambda_n,u_n)$ has norm bigger than 1. $\endgroup$
    – yolassr
    Sep 27, 2022 at 14:10
  • $\begingroup$ It has to be continuous in lambda as a consequence of the compactness of the operator. The question was answered in mathoverflow, look: mathoverflow.net/a/430932/173595 $\endgroup$ Sep 28, 2022 at 16:02

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