Estimate convolution error with non positive Kernel

Question

Let $f:\mathcal{C}^\infty(\mathbb{R})$. Let $\rho_n(x)$ be a family of convolution kernels, i.e. they have the following properties:

Have unitary mass: $\int\limits_{-\infty}^{+\infty} \rho_n(x) \text{d} x$ = 1,
Go to zero at infinity: $\lim\limits_{x \to \infty} \rho_n(x) = \lim\limits_{x \to -\infty} \rho_n(x) =0$,
$(\rho_n * f) \, (x) \to f(x)$ as $n\to \infty$. So, by identifying $\rho_n$ as a measure on $\mathbb{R}$, we have $\rho_n \rightharpoonup \delta_0$ (at least for the test function $f$).

Assume that $\rho_n$ has positive and negative values. Let $\mu_k(n) = \int x^k \rho_n(x) \text{d} x$ and $\tilde{\mu}_k(n) = \int x^k |\rho_n(x)| \text{d} x$ be the moments: assume that $\mu_k(n) \to 0$ as $n \to \infty$ with known rate, but $\tilde{\mu}_k(n) \not\to 0$.

Question: Is it possible to estimate the rate of convergence of the convolution $(\rho_n * f)(x)$, for example in function of $\mu_k(n)$ and $\tilde{\mu}_k(n)$?

That is, estimate $|(\rho_n * f)(x)-f(x)|$ with an explicit quantity that goes to zero as $n\to \infty$.

Background information and my attempts.

Let $\varphi(x)$ be a "nice" convolution kernel: a smooth mollifier, or a gaussian, and let $\varphi_n(x) = n \varphi(nx)$ be a rescaling which becomes thinner and higher. So $\varphi_n$ concentrates mass in point $x=0$, and therefore $\varphi_n \rightharpoonup \delta_0$. (I'm not being rigorous on what space we are using for weak convergence, but I talk about it only to give intuition; to be rigorous, think in terms of convolution kernels).

However, we know that concentrating mass is not the only way to converge in the weak sense. Other ways include mass translation that go to infinity, like $g + \mathbb{1}_{[n,n+1]} \rightharpoonup g$, or fast oscillations, like $g + \sin(nx) e^{-x^2} \rightharpoonup g$.

The family $\rho_n$ is of the second kind: $\rho_n$ is NOT a rescaling of some function $\rho$, but instead something that oscillates faster (and higher) as $n$ grows.

To estimate the rate of convergence for $\varphi_n$ instead of $\rho_n$, one can use Taylor series and regularity of $f$ to obtain a bound. Usually, the more regular $f$ is, the sharper the inequality. For example, following this approach we have:

$$ (f\ast \rho)(x) -f(x)=\int \bigl(f(x+ z)-f(x)\bigr)\rho(z) dz \approx \int \bigl(z f'(x) + \frac{1}{2} z^2 f''(x) + \ldots\bigr)\rho(z) \,\text{d}z. $$ Truncating at order 2 with Lagrange remainder, we obtain: $$ \left|(f\ast \rho)(x) -f(x)\right|\leq \left|f'(x) \int z\rho(z) \text{d}z\right| + \left|\frac{1}{2} \int z^2 \rho(z) f''(\xi(x)) \right| \text{d}z\leq $$ $$\leq |f'(x)| \left|\int z\rho(z) \text{d}z\right| + \frac{1}{2}\Vert f'' \Vert_\infty \int z^2 \left|\rho(z)\right| \text{d} z = |f'(x) \mu_1 | + \frac{1}{2}\Vert f'' \Vert_\infty \tilde{\mu}_2$$

If $\rho = \varphi_n$, then $\mu_1(n)=0$ and $\tilde{\mu}_2(n) = \frac{1}{n^2} \tilde{\mu}_2(0)$. So we obtain a bound $\left|(f\ast \varphi_n)(x) -f(x)\right| \leq \text{const} \frac{1}{n^2}$, which makes us pretty happy.

However, if $\rho = \rho_n$, then maybe $\mu_1(n)\to 0$ and $\mu_2(n) \to 0$, but $\tilde{\mu}_2(n)\not \to 0$, so our upper bound does not converge to 0.

Estimate convolution error with non positive Kernel

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