Let $$f(t,x)$$ such that $\lim_{(t,x)\rightarrow (0,0)}f(t,x)=f(0,0)$ and $$\int_0^1 f(0,0)dt=4f(0,0)$$ Then $$\int_0^1f(x,t)dt$$ is discontinous at $(0,0)$
(Previous one has error) Let $f(t,x)$ such that there is a cusp at $(1,0)$ and $\lim_{t\rightarrow 0}f(t,0)=f(0,0)=0$ and $\int_0^1 f(t,0)dt=f'(t,0)|_{t=1}$ Then $\int_0^1f(t,x)dt$ is discontinuous at $(1,0)$
@s.harp nah I am kinda studying the discussion about your question, because having a jump like integral discontinuity will suggest the function will have very interesting geoemetry at that point
and it is something I never heard before
Replace x with x-1/2, and you have negative bits popping up
(My question has been answered now, the previous answer was actually good enough) it turns out that there do exist continuous functions on $(0,1)^2$ so that the integral over one component results in a discontinuous function
@Secret are you missing some terms? That function is a product of two functions and thus doesn't have any behaviour different from a function on $\mathbb R$
(Desmos does not allow putting t, thus I put y whenever there is y to test). This one is kina hard to integrate, though, thus I am not sure if the graph agrees with the maths
If you are looking at just $y\cdot f(x)$ then the parameter $y$ will just give the amplitude of the how big $f$ is weighted, but it will always look like $f$, so you are not really gaining anything by looking at it as a function $\Bbb R^2\to \Bbb R$ as opposed to the function $f:\Bbb R\to\Bbb R$
I tried to use just the f(t,x) and use the geometry of the function to guess what I would get for $\int_0^1f(t,x)dt$ for some $x$. The answer in your question suggest functiosn that exhibit the behaviour you want will have those L shaped valleys
Of course, (if time) I can then use that to guide the actual math proofs on whether the function satisfy the required property
Or maybe I misunderstood, my brain is currently in knots now...
(sorry I would like to continue discussing and also draw some pictures of my own... but I really have to get back to correcting these execise sheets :P)
Well, @Danu, there are plenty of complex manifolds (even algebraic ones) with distinct complex (or algebraic) structures, even with tori. Or you can take a fixed manifold (say $\Bbb P^1$) and embed it in $\Bbb P^n$ in lots of different (non-isometric) ways (even different $n$) and look at the induced Kähler structures.
biholomorphic for sure ... I suppose you can say symplectomorphism for pulling back Kähler ... I was saying cohomology class as opposed to actual form, but ...
@TedShifrin I haven't actually come upon that terminology. Does it mean that the S-W classes are the ones pulled back from the taut. bundles over the infinite Grassmannians?
Those not in the know might end up searching for the meaning if you don't capitalize it. Capitalizing immediately shows it's derived from someone's name.
I used to have that book, @Alessandro. I recall there were a few mistakes in it. There's actually someone in the US who made an interactive webpage with that stuff.
Interestingly, there are a few flaws with LaTeX ... a few formatting commands (for lists, for example) which don't work the way they're supposed to. I figured that out by fiddling.
An interesting task is to write a journal with latex, where you just write the content of one entry in a file, give the file a name following some schema and then compile it all together in a master file
@GFauxPas Whether learning how to do things that people don't like is good or not? Yes a joke, but obviously its not the most important thing in the world that everybody thinks what you are doing is the correct way
it's just that whatever language I use other people say there's a better one, and I look at their languages and I think "I can do that in my language", whether it's Maple or R or something else
I found a nice way of finding Machin-like formulas. $\arctan\left(\frac{1}{17}\right)=\arctan\left(\frac{1}{27}\right)+\arctan\left(\frac{1}{46}\right)$
