@Akiva So a symplectic vector space is kinda like that, but your symmetric billinear form is now an antisymmetric billinear form (a 2-form). The right condition to replace "positive-definite" is non-degeneracy; f(u, v) = 0 for all u iff v = 0
First of all a sanity check, a compact operator between Banach spaces cannot be surjective, right? Because otherwise by the open mapping theorem we'd get an open precompact set in the codomain and that can't happen
if i have gradient for function $f(x,y)$ $$ \nabla f(x,y)=\begin{bmatrix} \frac{\delta}{\delta x} \\ \frac{\delta}{\delta y} \end{bmatrix} $$ so for example vector $ \hat{a} =\nabla f(x_0,y_0)$ points to the direction of largest increase of function $f(x,y)$ at point $(x_0,y_0)$. Now if i want to find out the largest decrease in $f(x,y)$ we can call this vector $\hat{b}$ is it just $\hat{b}= -\nabla f(x_0,y_0) $ we make both components negative ? that would flip vector in negative direction ?
but is inverse of $\hat{a}$=decreasing direction ? probably not ?
@Akiva The point of analogy of "positive definiteness" and "non-degeneracy" is the following. Suppose $(V, \langle - , - \rangle)$ is a (finite dim) inner product space; then the map $f : V \to V^*$ given by $f(x) = \langle x, - \rangle$ is an isomorphism. If $(V, \omega)$ is a symplectic vector space, similarly the map $f : V \to V^*$ given by $f(x) = \omega(x,-)$ is an isomorphism
@Liad Again, I'll add that I don't know measure theory. But here's a complete guess: isn't it because you have that $\nu(N) = 0$ and since it's a sum of two other measures, the other two measures must be $0$ as well because measures are nonnegative?
In the past month or two, the following answers are being quietly deleted two or three at a time without any explanation. I have spent significant amount of time and effort on these answers; I think I deserve some explanation.
I do refer to these answers from other forums once in a while; they a...
Actually, I vaguely remember someone who wanted to draw a huge picture of the Pythagorean theorem (the triangle with the squares on the sides) in the middle of the wilderness
"Assume aliens exist Then given that there are at least two intelligent races in the universe, humanity is not a unique event and there must be an infinite number of intelligent species in the infinite universe. At least one of these races must have discovered Pythagoras' theorem before Pythagoras. If that was the case then the theorem would have been named after the alien discoverer. BUT the theorem is known as pythagoras' theorem which is a contradiction. Thus aliens do not exist. QED"
My own position re: alien life is that the universe is both too big for alien life not to exist and too big for us to expect causal contact with alien life to occur
> "All of the science that we need to eventually create portals, stargate travel, hyperdimensional access mechanisms, levitation, teleportation, it all needs a mathematical foundation. And the closest that we have right now is the Ramanujan equations that we've been able to decipher so far, which came to him directly from this goddess who appeared to him in dreams."
@Semiclassical we can, for example, the greek and indians were far away and were not discovered till 16th century, but both had same mathematics, but in different languages, for ex- both invented pythogoras theorem,
So, even if there are aliens and we are isolated, we'd have a lot in common
> "Could Ramanujan have been telepathically receiving important information, that might enable humans to build a portal or stargate to other dimensions or alien worlds?"
What the hell happened to the History channel, man
if i have gradient for function $f(x,y)$ $$ \nabla f(x,y)=\begin{bmatrix} \frac{\delta}{\delta x} \\ \frac{\delta}{\delta y} \end{bmatrix} $$ so for example vector $ \hat{a} =\nabla f(x_0,y_0)$ points to the direction of largest increase of function $f(x,y)$ at point $(x_0,y_0)$. Now if i want to find out the largest decrease in $f(x,y)$ we can call this vector $\hat{b}$ is it just $\hat{b}= -\nabla f(x_0,y_0) $ we make both components negative ? that would flip vector in negative direction ?
I want to disprove the following statement.
Every Lipschitz-Continuous function is almost everywhere continuous differentiable.
From Whitney-Extension theorem, we know that the derivative of a Lipschitz-continuous function is continuous on a set being arbitrary close to a co-nullset. Does there...
@Semiclassical If you met Aliens, then how you'd explain to them what is called 'left' and what is 'right' to them without any physical contact in microphone, for example?
@Akiva so in effect theres an "antiderivative operator" $I$ that I can apply to both sides to "remove" the $D$ operator? $I(D(x^{2}y)) = x^{2}y = I(D(\frac{y^{2}}{2})) = \frac{{y^{2}}{2} + C$?
I want to disprove the following statement.
Every Lipschitz-Continuous function is almost everywhere continuous differentiable.
From Whitney-Extension theorem, we know that the derivative of a Lipschitz-continuous function is continuous on a set being arbitrary close to a co-nullset. Does there...
