# non compact closed range operator

Lately I've been reading Abramovich and Aliprantis' book 'An invitation to operator theory', chapter 2 (page 69) on bounded below operators. I would like to find an example of non-compact (and noninvertible) operator which is bounded below. It would be even better, if it were an integral operator. I know that

$$(Af)(x) := \int_{\mathbb{R}} \ e^{-|x-y|}f(y) \ dy$$

as an operator on $BC(\mathbb{R},\mathbb{R})$ (bounded and continuous functions) is non-compact. However, I cannot determine if its image is closed or not. I would really appreciate any help/hint.

## 1 Answer

For an operator that is bounded from below to be non-invertible, its range has to be a proper closed subspace of the codomain.

A simple example is the following operator on $L^p[0,1]$: $$(Af)(x) = \begin{cases}f(2x),\quad & x\in [0,1/2] \\ 0, & x\in (1/2,1] \end{cases}$$ Here $\|Af\|_p=2^{-1/p}\|f\|_p$, but the range of $A$ is clearly a proper subspace of $L^p[0,1]$.

You don't need non-compactness as a separate requirement. If an operator (on an infinite-dimensional space) is bounded from below, then it is non-compact.

Integral operators (such as convolution with $e^{-|x|}$ or anything similar) tend to not be bounded from below, because mollification erases small-scale features that contribute to the norm.