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Fixed the gap. Now the argument should be complete. Feel free to ask questions if something is unclear. The problem is equivalent to showing that the $L^4$ norm of a trigonometric polynomial $P(z)=\s …

$f(x)=\frac 12(1+\sin nx)$ is, indeed, the optimal choice. To see it, let's normalize a bit differently by $0\le f\le 2$. Then $f=1+g$ where $g$ is a real trigonometric polynomial of degree $n$ bounde …

The first maximum of $f(s)=\sin s+\sin\sqrt 2 s$ is attained at some $s_0\in [1.2,1.3]$ and exceeds $1.9$. The derivative of $f$ is at most $1+\sqrt 2\le\frac 52$. One obvious strategy is to go at the …

Indeed, I have just made things overly complicated in my approach. Thanks to JiK, I now see it fairly clearly. Let $(\{S\},P_S)$ be our probability space and let $X_k$ be the event $S\ni k$. Let $u_k …

For what it is worth, I played a bit with the non-constrained optimization of this type and noticed a strange thing. Suppose that you want to minimize $\sum_i a_i\max (F_i(x),0)^2$. Choose any $y$. In …

Unless you have made a mistake when typing the equations or are willing to abandon the non-decreasing property of $f$ and $g$ (and I'm not sure the latter will help), there is no solution. Indeed, mul …

What I would start with is the following: Let $x=\frac BNt$ be a new variable. Let's write the function as $$ G(x)=\Re[e^{-2\pi i Ax}P(x)],\quad P(x)=\sum_{k=1}^N r_ke^{-2\pi i (k-1-\frac N2)x} $$ wit …

The answer is "yes" and it is quite a nice linear algebra problem, but let me restate it first in a less intimidating way. We'll deal with $\mathbb R^n$ for any finite $n$. The first thing I will do i …

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 …

Here is a human verifiable proof. The quadratic equation for $x$ in terms of $a,t$ is $$ x^2-[(a-1)t+3]x+(2a-3)t=0 $$ Substituting $x=y+2$, we get $$ y^2+[1+(1-a)t]y=2+t=1+\frac{T-a}{1-a}\tag{*} $$ wh …

Not necessarily as written in the generality you want. Suppose that we are in $\mathbb R^1$, there is no linear constraint, and $Q(y)$ is some positive function of $y$. Then $v(y)\equiv 0$, which is c …

I'm not sure about stationary points, but the global minimum is certainly there. Let's do it for $n=5$. Write $$ f(x)=1+\frac{x_2}{2x_1}+\frac{x_2}{2x_1}+\frac{x_2x_4}{2x_1x_3}+\frac{x_2x_4}{2x_1x_3}\ …

This is not an answer but rather a nasty warning. Suppose you have 2 by 2 matrices and you fix the diagonal elements at some $q\in(0,1)$ and want to look at the out of diagonal elements that should be …

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 …

OK, sorry for the delay: yesterday was hectic and then I got distracted by another question. Here is a small piece of theory promised. First, let us notice that $\|L-SR\|_F=\|LR^*-S\|_F$ so, as you co …