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Apr
21
comment Existence of a continuous function which does not achieve a maximum.
A Little bit off Topic but is your $T_3$ necessarily Hausdorff?
Apr
21
revised Existence of a continuous function which does not achieve a maximum.
added 105 characters in body
Apr
21
answered Existence of a continuous function which does not achieve a maximum.
Apr
21
answered $L^p$ spaces and proper inclusion
Apr
21
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
Mar
27
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.
Mar
27
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
Feb
14
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Jan
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May
26
comment Show that $f(x)=e^x$
How did you define the Exponentialfunction? because often it is defined that way
May
22
comment Induced Matrix Norm
Because $\|A\|_1 = \sup_{x\in B_1(0)\setminus\{0\}} \frac{\|Ax\|_1}{\|x\|_1} \geq \frac{\|Ax\|_1}{\|x\|_1}$
May
22
comment Induced Matrix Norm
Well if you find such an $x$ you know that $\|A\|_1 \geq C$ and $\|A\|_1 \leq C$.
May
21
answered Convergent subsequence for sin(n)
May
15
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
May
15
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
May
15
answered the limit of a limit
May
15
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. $\lambda_1=\overline{\lambda_2}$
May
15
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?