I have an operator $A : L_{p}[0,1] \to L_{p}[0,1]$ which is $Ax(t)=\frac{1}{t}\int_{0}^{t}x(s)ds$ (namely Hardy operator) I want to show that this operator isn’t compact for $1 \leqslant p \leqslant \infty$ so for the case when $p=\infty$ I think this problem can be solved with constructing and example of sequence of functions for which from the image of this sequence we can’t deduce a fundamental sequence , but I can’t construct it. Also I want some hint for proving non-compactness for other cases
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A similar question has been asked before, it should help you work out some details: https://math.stackexchange.com/questions/2919027/t-l20-1-to-l20-1-tfx-frac1x-int-0x-fy-is-a-bounded-bu?noredirect=1&lq=1 – F. Conrad Mar 01 '20 at 12:00
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Yes, but still these method doesn’t work for $p=\infty$ – BeesaFangirl DOTO Mar 01 '20 at 12:39
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the construction in this answer works for $p=\infty$. – s.harp Mar 01 '20 at 12:55