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7h
comment Densely defined bounded integral transforms on $L^2(\Bbb R)$
@TrialAndError Oh yes I should say that for $f\in X$, the integral does exist in the Lebesgue sense, but it has the added property that $Tf$ is in $L^2(\Bbb R)$.
18h
comment Densely defined bounded integral transforms on $L^2(\Bbb R)$
That makes a lot of sense. The general case does seem pretty rough though. Using the fact that $L^2$ convergence implies pointwise (a.e.) convergence of a subsequence might be useful but it's not obvious to me where to go from there.
20h
asked Densely defined bounded integral transforms on $L^2(\Bbb R)$
1d
comment Difference between 1 (usual) and 1 bar of cayley table?
Because it's technically am equivalence class of numbers, not just a single number.
2d
comment Supervisor needs help. Is she really sick on Mondays?
She could be deathly allergic to Mondays. It's a serious condition that millions of Americans face every week.
2d
comment Since $\lim\limits_{x\to0}\frac{\sin kx}{kx}=1$ for constants $k$, is it also true for general arguments?
If $f$ is differentiable, l'Hospital's rule would give that $$\lim_{x\to 0}\frac{\sin f(x)}{f(x)} = \lim_{x\to 0}\frac{f'(x)\cos f(x)}{f'(x)} = \cos f(0) = 1.$$ For your purposes, this is a differentiable function so it works as expected. This assumes that $f'$ isn't zero in some neighborhood of $0$ though. If that is not the case, you might have to keep repeating the process.
2d
comment What is the limit of $x/(x+\sin x)$ as $x$ approaches infinity?
The tone of your post is pretty condescending.
2d
comment Integers with cubes ending $\ldots888$.
Please change your title. It's very unprofessional.
2d
comment We all use mathematical induction to prove results, but is there a proof of mathematical induction itself?
@MarioCarneiro The well-ordering principle is not the same as the well-ordering theorem. The well-ordering principle is often reserved for the naturals to avoid confusion (as the well-ordering principle is often brought up in the context of standard induction).
Aug
31
comment Why is the metric $d(f,g)=\int_a^b|f(x)-g(x)|dx$ important?
This is a continuum analogue of one of the basic metrics from real analysis: if $x,y\in\Bbb R^n$, define $d(x,y) = \sum_{i=1}^n |x_i - y_i|$. This is the taxicab metric.
Aug
31
answered Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$…$<\frac{n-1}{n}$
Aug
31
accepted The convergence of a product of sequences converging in $L^2$.
Aug
31
comment The convergence of a product of sequences converging in $L^2$.
Very true! It doesn't seem to be easily rectified. Oh well. This spurred from a consideration of the Fourier convolution theorem on $L^2$ (or its appropriate statement there, anyway). I was hoping to avoid the $L^1$ theory of the Fourier transform and stay within the realm of the $L^2$ theory but it seems not to be possible.
Aug
31
comment The convergence of a product of sequences converging in $L^2$.
This is a pretty good counterexample. I should have thought about reducing to the case of $f_n = g_n$. Coming up with separate sequences was becoming pretty painful. D'oh.
Aug
31
comment The convergence of a product of sequences converging in $L^2$.
The $L^4$ convergence seemed to be what I was circling around, but I was doubting myself on that front. I was considering that an $L^4$ convergence was necessary.
Aug
31
asked The convergence of a product of sequences converging in $L^2$.
Aug
30
comment Is this called an identity?
It's almost true. It's true for every value of $x$ except $x=a$.
Aug
30
comment about a maximal normal subgroup of a $p$ group.
You can't say that all of the normal subgroups are in the same chain. You will have in general different chains of inclusions going on. However each chain will necessarily terminate since $H$ is finite. Pick any one of those "final" normal subgroups and that is a maximal subgroup of $H$. Note that any of those will satisfy your definition since there will not be any other normal subgroups in $H$ which strictly contain them.
Aug
30
comment about a maximal normal subgroup of a $p$ group.
You can approach this from a "chain" perspective. Consider all of the normal subgroups contained in $H$. Then if you order these by inclusion, eventually you'll have to stop since $H$ is finite. These "final" subgroups are the maximal normal subgroups.
Aug
29
comment We all use mathematical induction to prove results, but is there a proof of mathematical induction itself?
It should be noted that induction can be proved, but only from an equivalent axiom (the well ordering principle). So really that it can be proved is equivalent to being an axiom anyway.