I am asking for pointers on how to move forward with the following proof, be that extended details regarding the current work or an alternative approach regarding the chosen mollifier and appropriate function sequence.
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a continuous function such that $\lim_{|x|\to\infty}f(x)=0$. I am trying to find a suitable mollifier $\phi$ so that $h_k\ast\phi_k$ is a compactly supported smooth function where $h_k$ is some suitable function possibly depending on $k$ and $\phi_k$ is a possible alteration of $\phi$ depending on $k$, such that $\lim_{k\to\infty}||f - h_k\ast\phi_k||_\infty = 0$ when $||f||_\infty \equiv \text{essential supremum}_{x\in\mathbb{R}^n}\{|f(x)|\}$.
To my knowledge mollifer is just some nice enough function used in convolution approximation, e.g. for $\phi\in C_0^\infty(\mathbb{R}^n)$, the function $\phi_k\equiv \frac{1}{k^n}\phi(x/k)$ is called the standard mollifier.
One idea I had was to take $\phi(x) = \exp\left(\left(|x|^2 - 1\right)^{-1}\right)$ when $|x| \leq 1$ and zero otherwise and $h_k(x) \equiv \chi_{\overline{B_k(0)}}(x)f(x)$. Then $h_k$, $\phi_k$ are compactly supported and $h_k\ast\phi_k$ is compactly supported smooth function.
What I am unsure about is how to then show that truly $\lim_{k\to\infty}||f - h_k\ast\phi_k||_\infty \equiv \text{essential supremum}_{x\in\mathbb{R}^n}\left\{\left|f(x) - \int_{\overline{B_k(0)}}f(y)\phi_k(x - y)dy\right|\right\} = 0$. Namely I am not sure how to force the entire expression to be less than $\epsilon$ on some closed ball $\overline{B_r(0)}$ so that the essential supremum "wants to live" outside of the said ball. I mean that since $f$ vanishes at the infinity, it follows that for any $\epsilon > 0$ there exists a closed ball $B_\delta(0)$ such that $|f(x)| < \epsilon$ outside of $B_\delta$. Hence if the said expression could always be forced to take $x$ further and further away from the origin, as $k$ increases, the limit would then be zero.