Squirtle
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 Jan 17 awarded Popular Question Dec 5 awarded Nice Question Nov 24 comment Applications of the Hahn Banach theorem for normed spaces? Yes, this works. This space is simple so your function that extends should be simple too, but it's a fair question. Nov 20 comment Proving two matrices are equal Your intuition for why it seems true is natural though and motivated by your familiarity with the $1\times 1$ real valued matrices. Nov 19 comment Complex integration and Gauss mean value theorem Thank you .... It was stupid easy and now I feel silly having asked. Nov 19 accepted Complex integration and Gauss mean value theorem Nov 19 comment Complex integration and Gauss mean value theorem If $|a|> 1$ it can be shown the integral you wrote is equal to $\log |a|$ by the Gauss MVT Nov 19 asked Complex integration and Gauss mean value theorem Nov 17 comment Computing the integral using cauchy's theorem If this is on the real line then you are just integrating at -1 and 1, so the integral is zero. I assume this is a complex integral? If so, it's more customary to use $z$ than $x$. Nov 16 comment Infinite dimension of a polynomial ring as a vector space Even the subspace $k[x_1]$ has infinite dimension since $1,x_1^1,x_1^2,\ldots$ is linearly independent. Nov 12 revised Riemann Lebesgue Lemma for locally compact ableian groups deleted 6 characters in body Nov 12 asked Riemann Lebesgue Lemma for locally compact ableian groups Nov 11 comment on the boundary of analytic functions How precisely is the dominated convergence used? I was thinking of using $g_n= \chi_{\Delta(\frac{n}{n+1})} g$ but it's not quite clear.... Nov 6 accepted Order of pole and evaluating residue Nov 5 awarded Popular Question Nov 2 awarded Notable Question Oct 29 revised Rationals are not locally compact and compactness added 12 characters in body Oct 28 comment How to prove this result involving the quotient maps and connectedness? Thank you! That makes sense. Oct 28 comment How to prove this result involving the quotient maps and connectedness? You wrote that $U=f^{-1}(f(U))$, this is true if $f$ is injective... however we are assuming $f$ is surjective (since it's a quotient map), the analogous result is that $U=f(f^{-1}(U))$....so this needs some cleaning up. Oct 27 awarded Popular Question