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4 votes
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Compact + absolute convergence of eigenvalues $\Rightarrow$ trace class?

Let $T: \ell^2\to \ell^2$ be defined by $$T(x_0,x_1,\ldots )=(0,a_0x_0,a_1x_1,\ldots )$$ where $a_n\to 0$ and $\sum |a_n|^2=\infty.$ Then $T$ is compact. Moreover $T$ does not admit any eigenvalues ...
Ryszard Szwarc's user avatar
2 votes

Does this inequality holds $ \int_{[0,1]} (f')^2 e^f dx \geq C \int_{[0,1]} f dx$ for some constant $C>0$?

Let $M=\int_0^1{(f’)^2e^f}$. By Cauchy-Schwarz and your assumption, $$M=\left(\int_0^1{(f’)^2e^f}\right)\left(\int_0^1{e^{-f}}\right) \geq \left(\int_0^1{|f’|}\right)^2.$$ Since $\int_0^1{e^{-f}}=1$ ...
Aphelli's user avatar
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