# Example of a compact operator that is not uniformly continuous

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.

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 $$B_1(0)$$.

• 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. Sep 21, 2022 at 15:03