An integrable C² density with bounded Hessian is bounded

Set $C=\max (1,H)$. For $x\in {\mathbb{R}}^d$, write $M=f(x)$ and $g=\mathrm{\nabla }f(x)$. For $q(t)=f(x+ty)$, the chain rule [F4] gives $q^{\mathrm{\prime }}(0)=g\cdot y$ and $q^{\mathrm{\prime }\mathrm{\prime }}(t)=y^T{D}^{2}f(x+ty)y$. Integration by parts and the second fundamental theorem [F5] give $q(1)-q(0)-q^{\mathrm{\prime }}(0)={\int }_0^1(1-t)q^{\mathrm{\prime }\mathrm{\prime }}(t)dt\mathrm{.}$ Since $\mathrm{\mid }{q}^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{\mid }\le C\mathrm{\parallel }y{\mathrm{\parallel }}^2$ and ${\int }_0^1(1-t)dt=1\mathrm{/}2$, this yields $\mid f(x+y)-M-g\cdot y\mid \le \frac{C}{2}\mathrm{\parallel }y{\mathrm{\parallel }}^2,$ so both the upper and lower Taylor bounds with this remainder hold. Taking $y=-g\mathrm{/}C$ and using $f\ge 0$ yields $0\le f(x-g\mathrm{/}C)\le M-\mathrm{\parallel }g{\mathrm{\parallel }}^2\mathrm{/}(2C)$; hence $\mathrm{\parallel }g\mathrm{\parallel }\le \sqrt{2CM}$.

If $M>0$, put $r=\frac{1}{8}\sqrt{M\mathrm{/}C}$. For every $y$ with $\mathrm{\parallel }y\mathrm{\parallel }\le r$, the corresponding lower Taylor bound and step 1.1 give $f(x+y)\ge M-\mathrm{\parallel }g\mathrm{\parallel }r-\frac{C}{2}{r}^2\ge M(1-\frac{\sqrt{2}}{8}-\frac{1}{128})\ge \frac{1}{2}M\mathrm{.}$ The cube $Q=x+[-r\mathrm{/}\sqrt{d},r\mathrm{/}\sqrt{d}{]}^d$ lies in that ball and has measure ${\lambda }_d(Q)=(2r\mathrm{/}\sqrt{d}{)}^d=(M\mathrm{/}(16Cd){)}^{d\mathrm{/}2}$ by [A1] and [F2].

Monotonicity [F3] and $\int f=1$ now imply $1\ge {\int }_{Q}fd{\lambda }_d\ge \frac{1}{2}M{\left(\frac{M}{16Cd}\right)}^{d\mathrm{/}2}\mathrm{.}$ Thus $M^{1+d\mathrm{/}2}\le 2(16Cd{)}^{d\mathrm{/}2}$ for every $x$ with $f(x)>0$; points where $f(x)=0$ satisfy the same resulting finite bound. Therefore $f$ is bounded.