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Jan
26
awarded  Yearling
Dec
14
awarded  Revival
Nov
19
answered Convergence of measures and potential theory
Nov
19
comment Covering map of the annulus
What have you tried?
Nov
18
revised Which functions have a list for all periodic points of them?
added 30 characters in body
Nov
18
answered Which functions have a list for all periodic points of them?
Nov
18
comment Integral under Diffeomorphsim, sign of Jacobian
That is another way to say it, yes!
Nov
18
answered Integral under Diffeomorphsim, sign of Jacobian
Nov
15
comment Equation for plane perpendicular to curve tangent hint
Yes, that is correct.
Nov
15
comment Lebesgue measure of proper subset
What happens to the measure of a closed interval if you remove one of its endpoints?
Nov
15
answered Equation for plane perpendicular to curve tangent hint
Nov
12
answered Gradient of a function twice
Nov
10
comment Maximization via Lagrange multipliers vs. substitution and partial derivatives
I'm not advising one over the other. However, in your comment, there is what sounds like a mathematical error: one must consider the possibility which maximizing a function on $z^2 + y^2\leq 5$ that that maximum lies on the boundary $z^2 + y^2 = 5$. If that were the case (it's not for this particular example but could be in another example) the maximum may not appear as a local maximum of the unconstrained function.
Nov
9
comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?
I think the idea of "outside" can really be subtle. For instance, if your surface is not orientable (like a Mobius band), then the word "outside" really has no meaning. But even for something like the triangle in your pictures: why is the normal vector you've drawn the "outside" one? Or imagine two cubes, one whose top face is a square in the $xy$-plane and another of the same size sitting on top of the first, so that its bottom face is the same square in the $xy$-plane. For these cubes, the outer normal vectors will be pointing different directions on this square.
Nov
9
comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?
Assuming it is orientable (so for instance, not a Mobius band), a surface in $\mathbb{R}^3$ always has a continuously varying (even smoothly varying) unit normal vector field. As for the notion of "outside" this only makes sense if the surface you're considering is closed. For instance, the sphere is closed, so it makes sense to talk about an outward unit normal, but the triangle in your pictures is not closed, so it is a choice to call the arrow you've drawn "outward" while the opposite is "inward".
Nov
9
answered If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?
Nov
9
comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?
If $M$ is a $2$ dimensional manifold, then its tangent space at any point has dimension $2$. If $M$ is embedded in $\mathbb{R}^3$, that means the dimension of the tangent space plus the dimension of the normal space must be $3$, so the dimension of the normal space is $1$.
Nov
9
comment Polar set of convex cones Proof
Awesome! No problem.
Nov
9
comment If $M$ is a submanifold of $\mathbb R^3$ and the normal space $N$ on $M$ at $p$ is one-dimensional, can we choose an unique “outer” normal from $N$?
It seems like you've given a pretty good answer already: if you can write your manifold $M$ as $\{\psi =0\}$ for some sufficiently regular function $\psi$, then $\pm \nabla\psi/\|\nabla\psi\|$ give the unit normal vectors (compactness of $M$ is not necessary). Another standard way of producing normal vectors is with parameterizations. That is, if $U\subseteq \mathbb{R}^2$ is open, and $f\colon U\to M\subset\mathbb{R}^3$ is a diffeomorphism onto an open subset of $M$, then the cross product of the partial derivatives $f_x(p)\times f_y(p)$ is normal to $M$ at $f(p)$ for all $p\in U$.
Nov
9
comment linear function, operator norm
What have you tried?