Schweitzer 2020

Let $T>0$ be a period of $f$. Work on the torus $\mathbb T=\mathbb R/\mathbb Z\times\mathbb R/T\mathbb Z$, with the homomorphism $\iota(x)=(x\bmod1,\ x\bmod T)$ from $\mathbb R$. The map $$h(u,v)=u+f(v)\bmod1$$ is well defined and continuous on $\mathbb T$, and $n\alpha+f(n\alpha)\equiv h(\iota(n\alpha))\pmod1$.

Reduce to a subgroup. Let $K$ be the closure of $\{\iota(n\alpha):n\ge1\}$. It is the closure of the cyclic subgroup generated by $\iota(\alpha)$ (in a compact group, the positive multiples already accumulate at $0$), so $K$ is a closed subgroup of $\mathbb T$. The points $h(\iota(n\alpha))$ are dense in the compact set $h(K)$. So it is enough to show $h(K)=\mathbb R/\mathbb Z$.

Use the structure of $K$. As a closed subgroup of a torus, $K$ is a compact Lie group: it has finitely many components, and its identity component $K_0$ is either a point, all of $\mathbb T$, or a closed circle subgroup $\{(pt\bmod1,\ qTt\bmod T):t\in\mathbb R\}$ for integers $p,q$ not both $0$. The first coordinates of $K$ form a closed set containing all $n\alpha\bmod1$, which are dense since $\alpha$ is irrational; so they cover the whole circle.

• $K_0$ cannot be a point: then $K$ would be finite.
• If $K_0=\mathbb T$, then $h(K)\supseteq h(\mathbb R/\mathbb Z\times\{0\})$, which is the whole circle.
• If $K_0$ is a circle with $p=0$, then $K_0$ has first coordinate $0$, and $K$, a finite union of translates of $K_0$, would have only finitely many first coordinates. Impossible.
• If $K_0$ is a circle with $p\ne0$, follow it for $t\in[0,1]$. The real function $\phi(t)=pt+f(qTt)$ is continuous and lifts $h$ along $K_0$, and $\phi(1)-\phi(0)=p+f(qT)-f(0)=p$. Since $|p|\ge1$, $\phi$ covers an interval of length at least $1$, so $h(K_0)$ is the whole circle.

In every possible case $h(K)=\mathbb R/\mathbb Z$, so the sequence is dense modulo $1$.

Step 1: convexity. For $p>1$ the function $u\mapsto|u|^p$ on $\mathbb R^n$ is convex and differentiable, with gradient $p|u|^{p-2}u$ (which is $0$ at $u=0$). So for every $x$, $$|x+Y(x)|^p\ \ge\ |x|^p+p\,|x|^{p-2}\,x\cdot Y(x).$$

Step 2: the extra term vanishes on balls. Let $H(\rho)=\int_0^\rho h(r)\,r^{p-1}\,dr$ and $\Phi(x)=H(|x|)$. Then $\Phi$ is $C^1$ with $\nabla\Phi(x)=h(|x|)\,|x|^{p-2}x$, which is continuous at $0$ because $p>1$. On the ball $B_\rho$ of radius $\rho$, since $\operatorname{div}(\Phi Y)=\nabla\Phi\cdot Y+\Phi\operatorname{div}Y=\nabla\Phi\cdot Y$, the divergence theorem gives $$\int_{B_\rho}\nabla\Phi\cdot Y\,dx=\int_{\partial B_\rho}\Phi\,Y\cdot\nu\,dS=H(\rho)\int_{\partial B_\rho}Y\cdot\nu\,dS=H(\rho)\int_{B_\rho}\operatorname{div}Y\,dx=0 .$$ The key point is that $\Phi$ is constant on each sphere, so it comes out of the flux integral, and the flux of a divergence-free field through a sphere is zero.

Step 3: integrate. Multiply Step 1 by $h(|x|)\ge0$ and integrate over $B_\rho$. By Step 2 the cross term drops out: $$\int_{B_\rho}h(|x|)\,|x|^p\,dx\ \le\ \int_{B_\rho}h(|x|)\,|x+Y(x)|^p\,dx .$$ Both integrands are nonnegative, so letting $\rho\to\infty$ (monotone convergence) gives the inequality over all of $\mathbb R^n$, with the usual convention if a side is infinite.