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May
6
comment The Analysis of Linear Partial Differential Operators I Prerequisites
To the poster: if you are a graduate student, asking a faculty member familiar with your background is probably better. Also, presumably as a graduate student in mathematics you have taken previously some undergraduate courses in mathematical analysis. Why not start by reviewing those?
May
6
comment The Analysis of Linear Partial Differential Operators I Prerequisites
@Ian: the scope and focus of Evans is quite different from Hormander; they cannot serve to replace each other. Otherwise I agree with your post. Rudin is also a good place to start for the OP. If the OP is also weak in differential calculus as is indicated, then one can even start with Rudin's Principles of mathematical analysis to get a quick review. Though I admit that in that situation it may be a year or so before the OP is ready to tackle Hormander.
May
6
revised The Analysis of Linear Partial Differential Operators I Prerequisites
edited tags
May
6
answered How does a Dirac delta function operate on a Fourier-Stieltjes integral?
May
5
comment Prove that $\lim_{x \rightarrow 0} \mathrm {sgn} \sin (\frac{1}{x})$ does not exist.
In fact, I am rather surprised that you were asked to show that this particular function has no limit at zero. If I were your instructor I would have assigned the two part question where the first part is to prove that for the function $f(x) = x \sin(1/x)$, the limit $\lim_{x\to 0^+} f(x)$ exists. And the second part to prove that $\lim_{x\to 0^+} \mathrm{sgn}\circ f(x)$ does not exist. This way you can see the difference that the signum function makes.
May
5
comment Prove that $\lim_{x \rightarrow 0} \mathrm {sgn} \sin (\frac{1}{x})$ does not exist.
@Diya: the signum function $\mathrm{sgn}$ returns $+1$ if its argument is positive, and $-1$ if the argument is negative. In particular: $\mathrm{sgn}(+1) = +1$ and $\mathrm{sgn}(-1) = -1$. So for the sequences $(x_n)$ and $(y_n)$ that you constructed above, you have $\mathrm{sgn} \sin(x_n) = \sin(x_n) = +1$ and similarly for $y_n$.
May
5
answered Reduce PDE to ODE
May
5
comment Prove that $\lim_{x \rightarrow 0} \mathrm {sgn} \sin (\frac{1}{x})$ does not exist.
@Diya: That looks fine.
May
5
comment How does a Dirac delta function operate on a Fourier-Stieltjes integral?
(1) As written, I don't think the reasoning is valid. (2) Yes, if you consider the fact that the Dirac delta function is not a function and $\delta(x) f(x)$ is not Riemann integrable. On the other hand, to actually conclude that the mean converges to "$dA(0)$" you don't need to reason "via" the delta function. You can do it rigorously and directly (at least when $dA(\omega)$ is replaced by $f d\omega$ where $d\omega$ is the Lebesgue measure and $f$ is some continuous function.)
May
5
answered Prove that $\lim_{x \rightarrow 0} \mathrm {sgn} \sin (\frac{1}{x})$ does not exist.
May
5
comment Schrödinger Kernels on manifolds
@AlexM. I don't know. It is something you should ask about on MathOverflow.
May
4
comment Finding the characteristic timescale of a first-order nonlinear ODE
the characteristic time-scale would be $1/a$. But if $c < a$ the characteristic time scale would be $1/c$. If $F(t) - 1$ is not exponential, but decaying like a polynomial, then $X$ does not decay exponentially at all, and so you cannot meaningfully speak of the characteristic timescale. In other words, the exact behaviour of other components, or in your case, the expected behaviour of the product $Y(t) Z(t)$ plays an important role in the asymptotics of $X$.
May
4
comment Finding the characteristic timescale of a first-order nonlinear ODE
Just to address your comment: in general it is not sufficient to just look at your $X$. Just look at the system $\dot{X} + aX = b F(t)$ where $F$ is some function of $t$ independent of $X$. And suppose that $F(t) \to 1$ as $t\to+\infty$. The actual characteristic timescale of $X$ depends on how fast $F$ converges. The equation above is equivalently $[e^{at} (X - b/a)]' = b e^{at} [F(t) - 1]$. Now suppose $F(t) - 1 = e^{-ct}$. If $c > a$
May
4
answered Finding the characteristic timescale of a first-order nonlinear ODE
May
4
comment Finding the characteristic timescale of a first-order nonlinear ODE
The definitions you quoted presupposes that the asymptotic behaviour of the system is that it will converge to a unique stable fixed point. For nonlinear systems this is no longer the case. Your system may even be unstable and have no fixed points.
May
4
reviewed Close Definite Integration problem
May
4
comment Curvature and convexity of a plane curve
the convexity of $\Omega$. Restricted to these curves what you want is actually true. See math.stackexchange.com/q/145808/1543 for a much more detailed discussion.
May
4
comment Curvature and convexity of a plane curve
When you enter the geometric picture and think about curves in the plane, there is no longer a canonical notion of "up", nor a global one at that. (You can say that "up" is in the direction of the normal vector, but that changes from point to point.) So a "convex curve" is not a good notion, at least from your definition. A better notion is to not treat arbitrary curves, but only curves that bound a set. Namely: let $\Omega$ be a subset of $\mathbb{R}^2$ with $C^2$ boundary $\partial\Omega$. Then you can ask about properties of $\partial\Omega$ as an embedded curve, and how it relates to
May
4
comment Curvature and convexity of a plane curve
@Bogdan: from the point of view of convex analysis, the problem is more that the notion of a "convex curve" is not a good notion. The correct notion of convexity is that "given two points in a convex set $C$, then any point 'in between' those two points are also in the convex set." When we say a function (or its graph) is convex, what we are saying is that the epigraph of the function is a convex set. And we can speak about the epigraph because when we are treating $\mathbb{R}$ valued functions there is a canonical notion of "up" using the ordering on $\mathbb{R}$.
May
4
revised Curvature and convexity of a plane curve
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