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It's a rare situation when the nature of equation makes an honest computation beat a smart consideration, when you shouldn't contemplate, only differentiate... In other words that is a case when to sh …

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 …

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 …

Zachary Chase said that he is still interested, so I'll make a post. It is a bit on the long side and I'll need to draw a few pictures to make it comprehensible, so today I'll just post a counterexamp …

Just take your favorite decreasing sequence $\lambda_k$ of positive numbers with sum $1$ and inductively construct the vectors $x_k\in C$ such that $\left \|\sum_{k=1}^n\lambda_k x_k-x\right\|\le\lamb …

Anyway, If you know a 5 line proof for the first inequality please share it with us OK, here goes. Assume that $\inf_P f=-1$ and that it is (nearly) attained at the origin. Let $K=\{x\in P: f(x)\le …

I would try to approach your original problem a bit differently. Note that $|x-a|\le \frac 12(r|x-a|^2+r^{-1})$ and the equality is attained for $r=|x-a|^{-1}$. Thus, $$ \min_x[|x-a|+\langle b,x\rang …