In my notes we stated that the following operator is bounded but not compact. Can someone give a short proof of it?
$T:L^2([0,1]) \rightarrow L^2([0,1])$ given by $(Tf)(x) = \frac{1}{x+1}f(x)$
In my notes we stated that the following operator is bounded but not compact. Can someone give a short proof of it?
$T:L^2([0,1]) \rightarrow L^2([0,1])$ given by $(Tf)(x) = \frac{1}{x+1}f(x)$
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