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Dec
16
accepted Why is this unbounded linear functional also has a closed kernel?
Dec
15
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Dec
11
accepted If a linear functional is not bounded, then it has a non-closed kernel.
Dec
11
comment If a linear functional is not bounded, then it has a non-closed kernel.
Oh, I get it. It's because according to one of the definitions of unbounded linear operator, there exists $y_n$, with $\|y_n\|=1$ but $|f(y_n)|\to\infty$, now let $x_n=y_n/|f(y_n)|$ and we get what we want.
Dec
11
comment If a linear functional is not bounded, then it has a non-closed kernel.
Why is $V/\ker f$ a finite dimensional space?
Dec
11
comment If a linear functional is not bounded, then it has a non-closed kernel.
Thanks. And why exactly we can find such sequence $x_n$?
Dec
11
asked If a linear functional is not bounded, then it has a non-closed kernel.
Dec
11
comment The kernel of a continuous linear operator is a closed subspace?
How is showing an unbounded sequence exists in a set proving this set being closed? There are closed sets that contain unbounded sequences. Please explain.
Dec
11
comment Why is this unbounded linear functional also has a closed kernel?
It should be "you're", sorry.
Dec
11
comment Why is this unbounded linear functional also has a closed kernel?
@Romeo: Yeah, your right, I forgot it has to be finite dimensional. Thanks.
Dec
11
asked Why is this unbounded linear functional also has a closed kernel?
Nov
8
comment The measure of the diagonal of a unit square in an alternative measure.
Why there's not even a comment? Did I make anything confusing here? Please point it out if you find the question confusing.
Nov
8
revised The measure of the diagonal of a unit square in an alternative measure.
added 49 characters in body