Dominic Michaelis
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 Apr21 comment Existence of a continuous function which does not achieve a maximum. A Little bit off Topic but is your $T_3$ necessarily Hausdorff? Apr21 revised Existence of a continuous function which does not achieve a maximum. added 105 characters in body Apr21 answered Existence of a continuous function which does not achieve a maximum. Apr21 answered $L^p$ spaces and proper inclusion Apr21 comment $L^p$ spaces and proper inclusion Are you sure that this is the exercise you should solve? Because the Statement is clearly wrong, as functions which are in $L^p$ for small $p$ are functions whose Peaks aren't to big and for big $p$ are those functions which goes to Zero fast enough Mar27 comment convergence of $a_n$ knowing that $a_n^n$ converges @peter.petrov sorry got a flaw, even $1$ may be an accumulation Point of the sequence. Mar27 comment convergence of $a_n$ knowing that $a_n^n$ converges @peter.petrov as I pointed out in the comments in Surb answer, the convergence of $(a_n^n)_{n\in \mathbb{N}}$ with the given setting only grants you that every accumulation Point is strict smaller 1 Feb14 awarded Yearling Jan29 awarded Stellar Question Sep30 awarded Explainer Jul2 awarded Curious May26 comment Show that $f(x)=e^x$ How did you define the Exponentialfunction? because often it is defined that way May22 comment Induced Matrix Norm Because $\|A\|_1 = \sup_{x\in B_1(0)\setminus\{0\}} \frac{\|Ax\|_1}{\|x\|_1} \geq \frac{\|Ax\|_1}{\|x\|_1}$ May22 comment Induced Matrix Norm Well if you find such an $x$ you know that $\|A\|_1 \geq C$ and $\|A\|_1 \leq C$. May21 answered Convergent subsequence for sin(n) May15 comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$ @Clin yes this is your inequality in the case they commute. I just want to know if there is an easy arguement why this should be true May15 comment Solve a problem of convergence of integral Hint: The Surface of the open ball with radius $r$ in dimension $N$ is a constant times $r^{N-1}$, while the measure of the ball is a constant times $r^N$. So I guess you can show that if there is no such sequence the integral of the function can't be finite May15 answered the limit of a limit May15 comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$ At least if $A$ and $B$ commute and are diagonalizable we have on the lhs $(\lambda_1^2 + \lambda_1 \mu_1+\mu_1^2)\cdot (\lambda_2^2 + \lambda_2 \mu_2 + \mu_2^2)$, where the $\lambda,\mu$ are the eigenvalues, if they are real we are done with arithmetic geometric mean and in the complex case the eigenvalues must be complex conjugate s.t. $\lambda_1=\overline{\lambda_2}$ May15 comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$ With Mathematica you can surely bruteforce it into an inequality with 8 unknowns. If we assume $A$ and $B$ to commute is there a simple argument why the lhs is positive?