Recent content by kingwinner

How about this? limsup means the limit of the supremum of terms with greater and greater index. If the whole set is bounded above, then sup_{k>0} is finite, right? sup(k>0), sup(k>1), sup(k>2),... forms a decreasing sequence right? The limit is either well defined real number or -infinity, isn't...

Is it possible to prove "directly" that sequence {x_k} is bounded above => limsup x_k < ∞ ??

Does this imply that the sequence {x_k} is bounded above => limsup x_k < ∞ ? Are you using contrapositive?

I think I'm OK with the direction: limsup x_k < ∞ implies the sequence {x_k} is bounded above. k->∞ (becuase if the sequence is not bounded above, then limsup=∞) But is the converse direction also true? The sequence {x_k} is bounded above implies limsup x_k < ∞ ? k->∞

Claim (?): limsup xk < ∞ k->∞ IF AND ONLY IF the sequence {xk} is bounded above. Does anyone know if this is true or not? (note that the claim is "if and only if") If it is true, why? Thanks!

I believe cosh(u) is actually a strictly convex function of u, and so I claim that 1 ∫cosh(u(x))dx := G(u) is also strictly convex, is this a correct implication? 0 Thanks!

Can someone help me, please? The definitions of convex/strictly convex of functionals (function of a function) are as follows: Let C be the class of C1 functions on interval [0,1] satisfying u(0)=0=u(1). A functional F is convex if for all u,v in C, 0<a<1, we have F((1-a)u+av) ≤...

Homework Statement Let C be the class of C1 functions on interval [0,1] satisfying u(0)=0=u(1). Consider the functional F(u)= 1 ∫[(u')2 + 3u4 + cosh(u) + (x3-x)u] dx 0 (note: u is a function of x.) Analyse the functional F term by term. Decide for each term whether it is convex or...

Let M={x|Ax=c} and f(x)=(1/2)x o Qx - b o x. I think you're suggesting there is an eaiser way to do the proof and I should probably discard my method? So I think you're suggesting that f: M->R is a convex function. But how can this be rigorously proved? Now A is not necessarily...

0 = φ'(0) = x1 o Qx0 - x0 o Qx0 - b o (x1-x0) and 0 ≤ φ''(0) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0 => φ'(0)+φ''(0)≥0 I think from here I need to show that f(x1)-f(x0)≥0. (which is a contradiction) Will a Taylor expansion work? But the problem is I think this will introduce a...

OK, surely without the constriant the statement can be false. But we are given that the constraint IS to be satisfied in this situation, and asked to PROVE that local min => global min under these assumptions. So I struggled for another day...still can't get the proof to work. :frown...

I'm sorry, but I don't understand the point you're making in relation to this proof. So I've shown that 0 = φ'(0) = x1 o Qx0 - x0 o Qx0 - b o (x1-x0) and 0 ≤ φ''(0) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0 => φ'(0)+φ''(0)≥0 I think from here I need to show that f(x1)-f(X0)≥0...

By the way, I would like to clarify that A and Q are matrices and x is some point in Rd (in case you are confused with the question).

Am I on the right track by adding φ'(0)+φ''(0)≥0? If not, can someone give me some hints on what to do next? It does not seem to lead to a contradiction. Thanks.

Homework Statement Homework Equations Constrined optimzation The Attempt at a Solution ("o" means dot product) Let M={x|Ax=c} and f(x)=(1/2)x o Qx - b o x Suppose x0 is a local min point. Suppose, on the contrary, that x0 is NOT a global min point. Then there must exist a...