For a Schwartz function $\psi(x)=xe^{-x^2}$ define $\varphi(x):=\psi'(x)$ and consider a family of $L^1$-dilations of $\varphi$ given by: 
$$
\varphi\_t(x)=\frac{1}{t}\varphi(x/t), \qquad t>0.
$$
$\textbf{Question:}$ Is there a function $f\in L^\infty(\mathbb{R})$ such that 
\begin{equation}\label{eq:1}
\liminf\_{t\rightarrow 0^+} |\varphi\_t\ast f(0)|>0,
\end{equation}
where $\ast$ denotes a convolution on $\mathbb{R}$ ?


If not, does exist $f\in L^\infty(\mathbb{R})$ satisfying a weaker condition
\begin{equation}
\int\_0^1 |\varphi\_t\ast f(0)|\, \frac{dt}{t}=\infty ?
\end{equation}


I editted the question after Aleksei's comment.



Example of a bounded function whose mean-zero mollification diverges at a point
$\newcommand{\ph}{\varphi}$
$\newcommand{\eps}{\varepsilon}$
Let $a\_k$ be a very fast-growing sequence of integers (I think $a\_k = 2^{1000k}$ should be enough). Consider the function $f$ defined as
$$f(x) = \sum\_{k = 1}^\infty \chi\_{[a\_k, 2a\_k]}.$$


I claim that there is a constant $c > 0$ such that for $\frac{1}{200a\_n} \le t \le \frac{1}{100a\_n}$ we have $\ph\_t(x)\*f(0) \ge c$. This is clearly enough since integral of $\frac{1}{t}$ over each of these intervals is $\log(2)$. We have


$$\ph\_t(x)\*f(0) = \sum\_{k = 1}^{n-1}\int\_{ta\_k}^{2ta\_k}\ph(-x)dx +\int\_{ta\_n}^{2ta\_n}\ph(-x)dx +\sum\_{k = n+1}^\infty\int\_{ta\_k}^{2ta\_k}\ph(-x)dx.$$


For the first sum we can bound it by the integral $\int\_0^{2ta\_{n-1}}|\ph(-x)|dx$, which is at most $2Cta\_{k-1}$, where $C = \max\limits\_{x\in \mathbb{R}} \ph(x)$ which we can make smaller than any $\eps > 0$ if $\frac{a\_{n-1}}{a\_n}$ is small enough.


For the third sum again we can bound it by $\int\_{ta\_{n+1}}^\infty |\ph(-x)|dx$. Since $\int\_\mathbb{R} |\phi(s)|ds$ converges we can make it smaller than any $\eps$ as long as $ta\_{n+1}$ is big enough, that is $\frac{a\_{n+1}}{a\_n}$ is big enough.


Finally for the middle term we have $x$ is always at least $\frac{1}{200}$ and at least $\frac{1}{50}$. For these $x$ we have that $\phi(-x)$ is at least some constant $p > 0$ (can be seen from the direct computation although the whole argument essentially works for any nonzero, continuous $L^1$ function).


In total we have
$$\ph\_t(x)\*f(0) \ge ta\_np - 2\eps\ge \frac{p}{200} - 2\eps.$$


Choosing $\eps = \frac{p}{1000}$ finishes the story.


As I said in the beginning, exponential grows of $a\_k$ is enough for this argument to work so we can actually replace $\frac{1}{t}$ by any decreasing positive weight $w(t)$ with $\int\_0^1 w(t)dt = \infty$.


