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Questions on the calculus of variations, which deals with the optimization of functionals mostly defined on infinite dimensional spaces.
9
votes
Accepted
An optimization problem
I'll change the variable $y=\frac 12-x$ to make typing easier. Since, as Peter already observed, the condition $Q(0)=0$ is worthless and since $\frac 1{12}$ is just an additive constant, we are just t …
3
votes
Accepted
Minimiser of a certain functional
As it has been already noted in the comments, the minimizer doesn't need to be unique. However, it always exists. It is not terribly hard to show but it is not trivial either, so I wonder why the ques …
1
vote
Upper bound on the maxima of ratio of expectation of quantities under Gaussian measure
I guess exercise 4.4.5 from the Estimates of Integrals chapter (Section "Positive integrals") of Stewart's "Calculus" (parallel universe edition) may be helpful.
Let $\mu$ be a probability measure on …
3
votes
Uniqueness of critical points for Lipschitz perturbations of uniformly convex Hamiltonians
Your $\cos$ construction can be easily mimicked in the setting you are interested in as well. Let $d=1$.
Let $\mu=pe^{-V}$, $g(x)=A\cos(ax)$ where $p,V,A,a$ are to be chosen. We'll take $p$ and $V$ ev …
2
votes
Accepted
Minimizing the expectation of a functional of probability distribution subject to an entropy...
Here is the lower bound of $0.49$ for all $\alpha\ge 1$. Note that $\min_\pi F(\pi)$ is a non-decreasing function of $\alpha$, so it is enough to consider $\alpha=1$. Also, the truth is about $0.55$ f …
5
votes
Do these surfaces intersect?
The answer is "yes" though the argument is rather ad hoc and doesn't generalize to vectors in more general positions.
We have $6$ unit vectors $v_j$ out of which the first $4$ are pairwise orthogonal …
8
votes
Accepted
Lavrentiev phenomenon between $C^1$ and Lipschitz
Then trivially yes. Just take $F(y)=1+\sum_{q\in \mathbb Q}\frac{a_q}{|q-y|^2}$ with $a_q>0$ such that the series converges a.e. Then change all $+\infty$ values of $F$ to $1$. Now take $[a,b]=[-1,1]$ …