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.


1 Answer 1


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 $B_1(0)$.

  • 1
    $\begingroup$ Actually, thinking about this: you can do the same thing with your $c_0$ example, just replace your $K$ by $K(\lambda, (u_k)) = (2 \max_k (\lambda^{1/k} |u_k|), 0, 0, 0, \ldots)$ and you'd get something compact. $\endgroup$ Commented Sep 21, 2022 at 15:03

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