@TedShifrin Wait , is @Alessandro the only person expert in logic ? What about @Mauro ALLEGRANZA , @Dave L. Renfero , @Brian M scott , @user:21820 , @Peter smith , @Noah Schweber , @Rob Athan , .... , etc ?
I have seen them from asking question in MSE about logic.
I honestly do not get the point or the price of that exam. Why are you making me solve random elementary math problems and asking words which I will never use. All that for a price of 200 dollars
so, for the proposition above, where $|r| < 1$, what's limit definition I need to invoke to argue $\lim \limits_{n \to \infty} r^{n+1} = 0$? Do I just treat it as a sequence? I ask because I'm not used to having half my equation be a sequence and not the other half, and at no point does it seem necessary to invoke the epsilon-delta definition for functions
@user I don't think that user21820 would be pinged here; they haven't been here in a while (at least they haven't said anything recently, if they have)
I'm a bit confused in the manner it is approaching $0$, since continuous functions seem to me to do it differently than sequences, but maybe you're right and it isn't a problem (am thinking through your answer)
insofar that, intuitively, I understand it. but, isn't there a specific definition of the limit that is invoked here, that isn't the typical epsilon-delta one for functions?
I thought there was only one definition of a limit (the epsilon-delta one)
@EdwardEvans I meant that if someone sees my username on my screen, they'll still find it harder to find my by simply searching in the Users tab because there's so many people named user or userXXXXX
@robjohn How can one find out which $\alpha$ in $k_\alpha (u)=\frac{1\{u\in[0,\alpha]\}}{\alpha}$ would make the inequality $\int |k_{\alpha}^2(u)u|du \le \int k{\alpha}(u)|u|du$ true? $1\{\}$ denotes the indicator function.
So plugging in $k_\alpha (u)=\frac{1\{u\in[0,\alpha]\}}{\alpha}$ in $\int |k_{\alpha}^2(u)u|du \le \int k_{\alpha}(u)|u|du$, one obtains $$\int |\left(1\{u\in[0,\alpha]\}\right)^2u|du \le \int 1\{u\in[0,\alpha]\}|u|du$$
@porridgemathematics tom Dieck's Transformation Groups, he attributes it to Bourbaki without proof (and there is no proof in the Bourbaki reference either)
Extremely quick question, bounded linear maps are ones that take bounded sets into bounded (image) sets. For example, the condition $|Tx|_Y \leq |x|_X$ for $T: X\to Y$ implies this definition, but how does it go in the other way? How does taking bounded sets into bounded sets imply that inequality?
one small question, where is boundless assumption in our quick algebra? is it containing |x|_X <\infty and so we can define the unit vector x/|x|? I noticed in the trivial case when we take the |0| we can't do what did, but the inequality still holds.
one of my analysis classes met right after a homological algebra class. i sometimes read what was still on the board. and i took algebraic topology once, so i saw some of it in that setting.
but homological algebra as homological algebra, never.
i got right up to the point where you had to understand spectral sequences and then i was like, you know what, i'm never gonna do this so i might as well stop now.
a lot of stuff that's just about bounds goes through with sublinear operators. i saw a good talk from larry evans on this once. he was interested in PDE but i wasn't and still got a lot out of the talk.
i love seeing a talk where someone pulls the rug out from under you. 'none of these operators need to be bounded,' 'you only use linearity to get part (d) of the list of 10 things that are otherwise equivalent for a larger class of maps.' galaxy brain stuff.
oh you know what? when we say bounded set here, we are always referring back to the quantity less than some finite (real) number, so this is still R, so LUB applies and hence sup is finite
i just looked up my instructor. he left for finance immediately after his postdoc. typical.
we just made a three-line submission to a judge. the signature block was followed by about a 3/4 page list of all the attorneys on the case. i live in paradise.
