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Let $M: C([0,1]) \rightarrow C([0,1])$ be defined by $$ Mf(x) = f(x/2), \;\; x\in[0,1]$$

Is this operator compact? I have trouble using limit in operator norm of compact operator, or cauchy subsequences... How do one do this in function spaces? If you can not find a clever counter example that is...

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2 Answers 2

up vote 4 down vote accepted

It is enough to show that $M(\mathrm{Ball}_{C([0,1])}(0,1))$ contains sequence without convergent subsequence. The most common example of such sequences are sequences $\{x_n:n\in\mathbb{N}\}$ with the property $$ \exists c>0\qquad\forall m,n\in\mathbb{N}\qquad m\neq n\implies \Vert x_n-x_m\Vert>c\tag{1} $$ Now consider triangle-shaped peaks $\{f_n:n\in\mathbb{N}\}$ of height $1$ with bases $[2^{-n-1},2^{-n}]$. You can check that $\{M(f_n):n\in\mathbb{N}\}$ satisfy $(1)$. The rest is clear.

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In fact, $M$ is an isometry of the infinite-dimensional closed linear subspace $\{g \in C[0,1]: g(t) = g(1-t) \text{ for all } t\}$ onto $C[0,1]$. So it's very far from being compact.

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