Let $(K_\delta)_{\delta > 0}$ be a family of integrable functions so that there is a constant $C \in (0 ,\infty)$ such that $\int K_\delta = 1$, $\int | K_\delta | \leq C$ for every $\delta > 0$ and for every $\varepsilon > 0$ we have \begin{equation*} (*) \quad \underset{\delta \to 0+}{\lim} \int_{| x | \geq \varepsilon} | K_\delta(x) | dx = 0 \end{equation*} Let $p \in [1, \infty]$. Prove that for every $f \in L^p(\mathbb{R}^d)$ we have $f * K_\delta \to f$ in $L^p(\mathbb{R}^d)$ as $\delta \to 0+$.
My thoughts:
I solved it for $p\in [1, \infty)$, but I believe there is a problem with $p = \infty.$ So here is my attempt at a counterexample:
let $$f(x) = \begin{cases} -1 & \text{ if } x\in (-\infty, 0] \\ 1 & \text{ if } x\in (0, \infty) \\ \end{cases}$$ and $K_\delta$ a non-negative smooth even function. Then \begin{equation*} (*) \quad f * K_\delta(0) = \int_{\mathbb{R}} f(-t)K_\delta(t) dt = \int_{\mathbb{R}_{\leq 0}} K_\delta dt - \int_{\mathbb{R}_{> 0}} K_\delta dt = 0 \end{equation*} Using $(*),$ we have \begin{align*} f*K_\delta(0) - f(0) &= \int_\mathbb{R} (f(-t) + 1)K_\delta(t) \\ &= \int_\mathbb{R} f(-t)K_\delta(t) dt + \int_\mathbb{R} K_\delta(t) dt \\ &= 0 + 1 = 1 \quad \text{ by $(*).$} \end{align*} Thus, \begin{equation*} \| f*K_\delta - f \|_\infty \geq 1 \end{equation*} For an example, consider \begin{equation*} K_\delta(x) = \frac{1}{\pi} \frac{\delta}{x^2 + \delta^2} \quad \text{ for }x\in \mathbb{R} \end{equation*}
Any feed back is much appreciated.
I'm not actually sure if the example I give at the end fits. It was an example given in Stein's Real Analysis of a kernel.
Your example is on the right track, but not pathological enough.
The result is false for $p=\infty$. To see this, assume that $K$ is smooth and compactly supported, and let $(K_\delta)_{\delta>0}$ be the approximate identity generated by $K$. Then for any $f\in L^\infty(\mathbb{R})$, $K_\delta*f$ is a continuous function (in fact uniformly continuous). But let $E\subset[0,1]$ be a fat Cantor set (i.e. a Cantor set of positive measure), and let $f$ be the indicator function of $E$. Then $K_\delta*f$ cannot converge in $L^\infty$ to $f$. For each $K_\delta*f$ is continuous, and for continuous functions convergence in $L^\infty$ is equivalent to uniform convergence. But then $K_\delta*f$ must be a sequence of continuous functions converging uniformly, and hence the limit must be continuous. Call this limit $f_\infty$; then $\|f_\infty - f\|_\infty > 0$. For $f$ cannot be equal a.e. to a continuous function: such a continuous function would be $0$ on a dense subset of $[0,1]$, and therefore $0$ everywhere, but also $1$ on a positive measure subset of $[0,1]$, which is absurd.
What is true is that if $f\in L^\infty$, then $K_\delta*f$ will converge to $f$ on the set where $f$ is continuous. The convergence will be uniform on the set where $f$ is uniformly continuous.