Meital

Has anyone studied Differential Geometry from Do Carmo's book, Riemannian Geometry? I am trying to work problem 4 page 57, but I don't even know how to start. I will scan the problem and post it laer.

ok i figured it out!

ODE/Sturm-Liouville problem. consider the Sturm-Liouville problem: y" + [ lambda p (t) - q(t)] y= 0 in (0,1), alpha y(0) + beta y'(0) = 0 gamma y(1) + delta y'(1) = 0, Where alpha, beta, gamma, delta are real constants, and p: [0,1] -> R and q:[0,1] -> R are continuous functions with p(t) > 0. (a) Suppose alpha*beta doesn't equal 0. Show that if f_n(t) and g_n(t) are eigenfunctions associated with a given eigenvalue lambda_n of the Sturm-Liouville problem, then f_n(t) = c g_n(t), t belongs to [0,1], for some constant c in R. (b) Can one remove the restriction alpha*beta doesn't equal to 0 in part (a)? [ Explain and justify your answer.]

Prove or disprove: If the boundary of set omega in R^d has an outer measure zero, then omega is Lebesgue measurable. I was trying to come up with a counter example, but I couldn't. Then I tried to prove it, yet I was not able to do so. Please help me

If A is Lebesgue measurable and x is an element of Rn, then the translation of A by x, defined by A + x = {a + x : a ∈ A}, is also Lebesgue measurable and has the same measure as A.

Can you guys tell me if my answer is correct? Determine if the following functions satisfy local or uniform Lipschitz condition. 1). te^y I found d/dy (te^y) = te^y, and this is not bounded above for any value of y, so this made me conclude that it has locally Lipschitz condition since the Lipschitz constant here changes as the reagion changes? Am I right? I used the equation | f(t,y_1) - f(t, y_2) | = d/dy(f(t,y)) + | y_1 - y_2| 2). y t^2/ (1 + y^2) I used the same approach here and d/dy (f ) = t^2 - 2y + y^2/ ( 1 + y^2)^2, which is clearly could be bounded above by a constant but this constant changes as the reagion changes so it is local lipschitz.

I am reading ODE ( Ordinary DE) notes, and there is a statement that says " A function that is not uniform, even a continuous function that is not uniform, cannot have a lipschitz constant. As an example is the function 1/x on the open interval (0,1). I want to see if I understood this correctly. I will assume that we can find a number k >= 0 such that | f(x) - f(y) | =< K*|x - y| we have x and y in (0,1) so 0 < x < 1 0 < y < 1 .....(1) Now, | 1/x - 1/y | =< k*|x - y| Find common denominator | y - x|/|xy| =< k*|x-y| | y - x | = |x - y| then we cancel the term from both sides of inequality, so we get 1/|xy| =< k ...($) k >= 0, and from (1) 1/xy > 1 so is 1/|xy| > 1 but then the equation ( $ ) becomes 1 < 1/|xy| =< k , but k >= 0 so this is contradiction? I am not sure, can someone tell me why 1/x doesn't have a lipschitz constant? I mean since we are in (0,1) then there is no upper bound for 1/|xy| so we can't have an upper bound for it (lipschitz constant)

I figured it out..I don't know why I thought it was difficult! Thanks though

Can someone help me with the following proof: Suppose f ,g: X -> [- infinity, + infinity] are measurable. Prove that the sets { x : f(x) < g(x) } , {x: f(x) = g(x) } are measurable.

ok i figured it out

I am trying to find the set A such that For r > 0 let A ={w, w = exp (1/z) where 0<|z|<r}.

what is the radius of convergence of the sum (n=0 to infinity) of z^(n!) ?

ok I got it..no need for any replies..thanks though

Can someone find an example of a sequence such that it doesn't have a limit, but it has a lim sup? and find the lim sup? I thought about the sequence of all rational numbers in the interval [0,1], but not sure if that's a correct example.

I see, so we can take the colletion: R ( real numbers), open set (0,2), and the empty set. This is topology since it satisfies all 3 axioms of topology, but not sigma-algebra because we don't have the comp of (0,2) in the collection.