Example of a bounded function whose mean-zero mollification diverges at a point 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.
 A: $\newcommand{\ph}{\varphi}$
$\newcommand{\eps}{\varepsilon}$
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 $\int_0^1 w(t)dt = \infty$.
