Second derivative of mollification at local maximum Let $u:[-1,1] \mapsto \mathbf{R}$ be a continuous function with a local maximum at 0. That is, there is a ball $B_\delta(0)$ with $u(x) \leq u(0)$ for all $x \in B_\delta(0)$. Let $\eta(x)$ be the standard mollifier defined by
$$ \eta(x) = \begin{cases} C \exp\left(\frac{1}{|x|^2-1}\right) &\text{if } |x|< 1\\ 0 &\text{if } |x| \geq 1,\end{cases} $$
where $C$ is chosen so that $\int_{\mathbf{R}^n} \eta = 1$. For $x \in (-1+\epsilon, 1 - \epsilon)$ define the mollified function $u_\epsilon$ by
$$ u_\epsilon(x) = \int_{B(0,\epsilon)} \epsilon^{-n}\eta\left(\frac{y}{\epsilon}\right)u(x-y) \ dy. $$
Is it true that
$$ \liminf_{\epsilon \rightarrow 0} u_\epsilon ''(0) \leq 0. $$
This would be true if the mollified function also had a local maximum at 0 — however this is not necessarily the case. As some further intuition it should be true that the mollification has a local max near 0, and this local max approaches the origin as $\epsilon \rightarrow 0$. However without something resembling uniform convergence I've been unable to turn this into a solution.
Note also the lim inf can not be replaced by a limit as we can construct a function for  for which $ \lim_{\epsilon \rightarrow 0} u_\epsilon ''(0)$ does not exist. The idea behind the construction is to take any positive function for which the limit of the mollifications don't exist at 0 then integrate twice. More precisely take $h \in L^\infty[0,1]$ with $h \geq 0$ for which $\lim_{\epsilon \rightarrow 0} h_{\epsilon}(0)$ does not exist. Set $g(x) = \int_0^x h(x)$ and $u(x) = \int_0^x g(x)$. Note $u(0)=0$ and $u \leq 0$. Then $h$ bounded implies $u,g$ are Lipschitz, and differentiable a.e with $u′=g$ and $g′=h$. Since the derivative of the mollification is the mollification of the derivative we have $(f_\epsilon)'' = u_\epsilon$ which we recall was chosen to not have a limit at 0.
 A: So, this is an incomplete solution that hopefully can give some ideas. The spirit here is to find a counterexample to the main statement. What I have managed to do is to find a $u$ for which $\limsup_{\epsilon\to 0^+}u_\epsilon''(0)>0$, and I hope to manage to obtain the same thing for the $\liminf$ with a similar construction. The idea is to build $u$ such that $u_\epsilon''(0)$ oscillates with some sort of periodicity as
$\epsilon\to 0^+$.
I don't know if the statement is true or not though, but also trying to exhibit counterexamples could give ideas for proving that a statement is true.
I hope it is not a problem, but I feel more comfortable changing the domain of $u$ in $\mathbb{R}$, and assuming
$$u:\mathbb{R} \mapsto \mathbb{R}\;\;\text{continuous},\qquad u(0)=0\,,\qquad u(x)\leq 0\quad\forall x\in \mathbb{R}.\qquad \text{(HP)}$$
I think we agree that this is equivalent to the original problem.

Definition of self-similar function
Let's consider a function $u$ with the hypotheses (HP) and such that $\exists\, \delta>0,\alpha>1$ such that
$$u(x)=u(\alpha x)/\alpha^2\qquad\forall x\in[-\delta,\delta].$$
I will call it self-similar function, as it also holds from that that
$$u(x)=u(\alpha^n x)/\alpha^{2n}\qquad\forall x\in[-\delta/\alpha^{n-1},\delta/\alpha^{n-1}].$$
Let's compute $u_\epsilon''(0)$ assuming $\epsilon\leq\delta$:
$$u_\epsilon''(0) = \int_{B(0,\epsilon)} \epsilon^{-3}\eta''\left(\frac{y}{\epsilon}\right)u(y) \ dy=\int_{B(0,\epsilon)} \epsilon^{-3}\eta''\left(\frac{y}{\epsilon}\right)u(\alpha y)/\alpha^2 \ dy=$$
$$=\int_{B(0,\alpha\epsilon)} \epsilon^{-3}\eta''\left(\frac{z}{\alpha\epsilon}\right)u(z)/\alpha^3 \ dz=u_{\alpha\epsilon}''(0) $$

Let's make the change of variable $s:=-\ln(\epsilon)$, i.e. $\epsilon=\exp(-s)$. The above computation means that the function
$$ f_u(s):= u''_{\exp(-s)}(0) $$
is periodic with period $\ln \alpha$ in the interval $s\in[-\ln\delta,+\infty)$, as we have
$$ u''_{\exp(-s)}(0)=u''_{\alpha\exp(-s)}(0)=u''_{\exp(+\ln(\alpha)-s)}(0)\qquad\forall s\geq-\ln\delta \iff \epsilon\leq\delta.$$

Why a self-similar function can be useful to prove or disprove the statement
So, since $\epsilon\to 0^+$ is the same thing of $s\to+\infty$, we are interested in the behaviour of $f_u(s)$ for large $s$. In particular, our problem is equivalent to prove that for all $u$ that satisfy the hypotheses (HP), it holds
$$ \liminf_{s\to+\infty}f_u(s)\leq 0$$
For self-similar functions $u$, we have proved that $f_u(s)$ is periodic for $s$ large enough. If we could exhibit a self-similar function $u$ such that $f(s)\geq c>0$ for one whole period of $f$, i.e. for $s\in[s_0,s_0+\ln\alpha]$ where $s_0\geq-\ln\delta$ and for some $c>0$, we would find a counterexample and the claim is false.

A partial result, and an example of construction of a self-similar function
At the moment, it seems a bit complicated to find such $u$ if it exists, but I can actually exhibit an example of $u$ such that
$$ \limsup_{\epsilon \rightarrow 0} u_\epsilon ''(0) > 0, $$
that is,
$$ \limsup_{s\to+\infty}f_u(s)> 0.$$
Since $f_u$ is periodic for $s\geq-\ln \delta$, it just suffices to find $s_0\geq-\ln \delta$ such that $f_u(s_0)>0$.
My example is that of a function that is only continuous in $0$ for the sake of simplicity and has some jumps, but with some more calculations, one could build a function with the same properties that is also continuous (for example, regularizing $\phi$ before the following construction).
Let's thus consider
$$ \phi(x):= -\chi_{[-1,-\gamma]}(x)-\chi_{[\gamma,1]}(x),$$
where $\chi_I$ is the characteristic function of the interval $I$ and $0<\gamma<1$ is a small constant to be chosen later. Then let's define $u$ as follows:
$$ u(x)=\sum_{n=-2}^\infty \phi(\alpha^nx)/\alpha^{2n},$$
where $1<\alpha$ is to be chosen later, and we assume $\gamma^{-1}<\alpha$ so that the characteristic functions don't overlap.
You can check that $u$ is well-defined, is continuous in $x=0$, it is uniformly bounded by $0$ from above and it's self-similar with constants $\delta=\alpha$, and $\alpha$ is the same $\alpha$ in the definition of self-similarity. Rewriting the conditions that I have written in the beginning of this section, it's just necessary to prove
$$ \exists \;0<\epsilon<\alpha\quad\text{such that } u_\epsilon ''(0) > 0.$$
This is true if we choose:
(1) $\epsilon=2$
(2) $0<\gamma<1$
(3) $\alpha>4/\gamma$
Unfortunately, this is just a boring matter of estimating $u_\epsilon''(0)$, as the explicit computation would be very long. The idea is that choosing $\epsilon=2$ (that is, big but not too big), the term of the series for $n=0$ has positive second derivative in $x=0$ when regularized with parameter $\epsilon$, because of the facts that $-\eta$ has positive second derivative near $x=0$ and $\phi$ is negative. The terms with $n\geq 1$ have a positive contribution for the same reason. The contribution of the terms for $n=-1,-2$ is $0$ because $\alpha$ is taken very large, and the support of their regularization does not touch $x=0$.

So...
So, the question that I have now is: does this construction work to prove the same result for the $\liminf$? We still have some freedom in choosing $\gamma$, which I have basically not used, but it seems more difficult... I will try to think about it. I hope that someone has some clever idea for a simpler proof :)
