Prove or disprove: $e^{-nG(x)}$, normalized, is an approximation to the identity for $G(x)$ strictly convex We are given the sequence of functions
$$ \phi_{n} = \frac{e^{-nG(x)}}{\int_{\mathbb{R}}e^{-nG(x)}dx}$$
for a nonnegative, strictly convex function $G$ (that is, $G'' \geq c$ for some $c>0$) that attains its global minimum at 0.
It is clear that $\phi_{n} \geq 0$ and $\int_{\mathbb{R}}\phi_{n}=1$ for each $n$. For the $\phi_{n}$ to be considered an approximation to the convolutive identity (that is, they approximate a dirac delta function), we require that for each $\delta>0$ and $\epsilon>0$,
there exists an $N$ large enough that
$$ \int_{|x|>\delta}\phi_{n} < \epsilon$$
for each $n>N$.
Is it true that the $\phi_{n}$ satisfy this last property for $G$ as specified? I suspect it is true, due to known examples like $x^{2}$.
If true, can we loosen the conditions on $G$ to include other known functions like $|x|$? Perhaps a sufficient condition is that $G$ attains its global minimum exactly once, is strictly decreasing to the left of that minimum, and strictly increasing to the right of it.
 A: Let $G$ be convex and with minimum at zero — and only at zero.
(We don't need $G$ to be nonnegative, nor do we need it to be strictly convex.  Note that our assumptions allow $G(x)=|x|$, as requested at the end of the question.  In fact, the idea of this proof is essentially to use convexity to relate $G$ to a function made up of two straight lines, like $|x|$, and to prove it for such functions directly.)
Let $\delta>0$.  We wish to show $\int_\delta^\infty \phi_n(x)\,dx \to 0$ as $n\to\infty$, and the same for $\int_{-\infty}^{-\delta}$.  I'll do the first; the second is similar.  Consider the secant of the graph of $G$ over $[0,\delta]$:

The secant is $y=a+bx$, where $a=G(0)$ and $b=(G(\delta)-G(0))/\delta$.  (Note that $b>0$ because $G(0)$ is the only minimum of $G$.)  As the diagram suggests, since $G$ is convex,
\begin{align*}
G(x) &\le a+bx &&\text{if $0\le x\le\delta$,}\\
G(x) &\ge a+bx &&\text{if $\delta\le x$.}
\end{align*}
Then
\begin{align*}
\int_\delta^\infty e^{-nG(x)}\,dx
\le \int_\delta^\infty e^{-n(a+bx)}\,dx
= e^{-na} \int_\delta^\infty e^{-nbx}\,dx
= \frac{e^{-na}}{nb} \int_{\delta bn}^\infty e^{-u}\,du
\end{align*}
(where the substitution $u=nbx$ is legitimate because $b\ne 0$), and
\begin{align*}
\int_{-\infty}^\infty e^{-nG(x)}\,dx
\ge \int_0^\delta e^{-nG(x)}\,dx
\ge \int_0^\delta e^{-n(a+bx)}\,dx
= \frac{e^{-na}}{nb} \int_0^{\delta bn} e^{-u}\,du
\end{align*}
Dividing,
$$ \int_\delta^\infty \phi_n(x)\,dx
= \frac{\int_\delta^\infty e^{-nG(x)}\,dx}
  {\int_{-\infty}^\infty e^{-nG(x)}\,dx}
\le \frac{\int_{\delta bn}^\infty e^{-u}\,du}
  {\int_0^{\delta bn} e^{-u}\,du}
\to 0 \qquad\text{as $n\to\infty$.}
$$
(We used $b>0$ again at the end there, to make sure that $\delta bn$ increases with $n$.)
