# Example of a compact operator that is not uniformly continuous. [Solved]

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.

• The second condition also should mention $K$. As it is written now, it does not make sense.
– daw
Sep 19, 2022 at 6:08
• Thank you! It is correct now. Sep 19, 2022 at 13:57
• 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. Sep 27, 2022 at 14:10
• 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 Sep 28, 2022 at 16:02