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seen Jun 5 at 10:33

May
12
comment Continuous representation restricts to homomorphism $G \to O(n)$
Do you remember how this was done for G finite? Can we do a similar thing...?
May
4
comment How is the Fourier transform “linear”?
Indeed your standard linear functions of the form $f(x) = ax + b$ aren't actually linear at all when $b \ne 0$ (since linear functions map 0 to 0). These straight line functions comprise the more general class of affine transformations
Apr
19
comment Integral metric.
The space C[0,1] equipped with the above norm isn't a Hilbert space: it fails on both counts. It isn't complete, consider $f_m(x) = 0 $ for $0 \le x \le \frac{1}{2} - m $ and $f_m(x) = 1 $ for $ \frac{1}{2} + m \le x \le 1$ and f being a straight line segment joining 0 and 1 between these two intervals. It's a Cauchy sequence, but it would necessarily converge to a function with a discontinuity at x= 1/2. So we could consider its completion, but this space cannot be an inner product space since it does not obey the en.wikipedia.org/wiki/Parallelogram_law
Apr
19
answered Integral metric.
Apr
15
comment How to prove that the inverse exists on the whole space X?
A linear map is injective if and only if it has trivial kernel. So suppose there is some non-zero x with (I-T)x = 0. Then x = Tx so $||Tx||/||x|| \ge 1 \Rightarrow ||T|| \ge 1$, a contradiction.
Apr
13
answered Evaluation of $\sum_{x=1}^{\infty}x^{-x}$
Apr
13
answered Solution set of cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x))))
Apr
12
awarded  Supporter
Apr
12
awarded  Student
Apr
12
asked Cutesy Applications of Fermat's Last Theorem (or others)
Apr
12
awarded  Teacher
Apr
12
answered Mapping a variable having very vast range to the interval (0,1)