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Jan 13 at 3:26 vote accept luyao
Jan 12 at 19:17 comment added Iosif Pinelis @GiorgioMetafune : Thank you for this nice comment.
Jan 12 at 18:40 comment added Giorgio Metafune This is another proof: if $T$ is positive, $(Tx,x) \leq c (x,x) $ is equivalent to $\|T^{1/2}x\|^2 \leq c\|x\|^2$ or $\|y\|^2 \leq c \|T^{-1/2}y\|^2$ which is $(T^{-1}y,y) \geq c^{-1}(y,y)$. In the general case $0 \leq T \leq S$ is equivalent to $S^{-1/2} TS^{-1/2} \leq I$ which, by the previous case, gives $S^{1/2} T^{-1}S^{1/2} \geq I$ or $T^{-1} \geq S^{-1}$.
Jan 12 at 17:37 history edited Iosif Pinelis CC BY-SA 4.0
added 320 characters in body
Jan 12 at 17:14 history answered Iosif Pinelis CC BY-SA 4.0