# Proof that Good Kernels are Approximations of Identity in $L^p(\mathbb R^d)$

$$\textbf{The Problem:}$$ Suppose that $$(K_{\delta})_{\delta>0}$$ is a family of integrable functions such that there exists a constant $$C\in(0,\infty)$$ such that $$\int K_{\delta}=1,\int\vert K_{\delta}\vert\leq C$$ for every $$\delta>0$$ and for every $$\eta>0$$ we have $$\color{blue}{\large\lim\limits_{\delta\to0^{+}}\int_{\vert x\vert\geq\eta}\vert K_{\delta}(x)\vert dx=0}.$$ Let $$p\in[1,\infty)$$. Prove that for every $$f\in L^p(\mathbb R^d)$$ we have $$f\ast K_{\delta}\to f$$ in $$L^p(\mathbb R^d)$$ as $$\delta\to0^{+}.$$

$$\textbf{My Thoughts:}$$ Here we go. For $$p\in[1,\infty)$$ we use Minkowski's inequality for integrals, for reference, this is $$6.19$$ on page $$194$$ of Folland's Real Analysis, $$2$$nd Edition. With this in mind we have, let $$\varepsilon>0$$ be given, then we have that there is $$\eta>0$$ such that $$\|\tau_yf-f\|_{p}<\varepsilon$$ for all $$\vert y\vert<\eta.$$ Putting these together we have \begin{align*}\large\|f\ast K_\delta-f\|_p&=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(x-y)K_\delta(y)dy-f(x)\Bigg\vert^p dx\right)^{1/p}\\ &=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(x-y)K_\delta(y)dy-\int_{\mathbb R^d}f(x)K_\delta(y)dy\Bigg\vert ^p dx\right)^{1/p}\\ &=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}[f(x-y)-f(x)]K_\delta(y)dy\Bigg\vert ^p dx\right)^{1/p}\\ &\leq\large\int_{\mathbb R^d}\left(\int_{\mathbb R^d}\vert f(x-y)-f(x)\vert^{p}dx\right)^{1/p}\vert K_{\delta}(y)\vert dy\\ &\leq\large\int_{\mathbb R^d}\|\tau_{y}f-f\|_{p}\vert K_{\delta}(y)\vert dy\\ &\leq\large\int_{\vert y\vert\geq\eta}2\|f\|_{p}\vert K_{\delta}(y)\vert dy+\int_{\vert y\vert<\eta}\|\tau_{y}f-f\|_{p}\vert K_\delta(y)\vert dy\\ &\leq\large2\|f\|_{p}\int_{\vert y\vert\geq\eta}\vert K_{\delta}(y)\vert dy+\varepsilon\\ &\large\overset{\delta\to0^{+}}{\longrightarrow\varepsilon.} \end{align*} It follows that $$f\ast K_{\delta}\to f$$ in $$L^p(\mathbb R^d)$$ as $$\delta\to0^{+}.$$

Do you agree with the proof presented above?

Any feedback is much appreciated.

As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $$(X,\frak{M},\mu)$$ and $$(Y,\frak{N},\nu)$$ are $$\sigma$$-finite measure spaces, and let $$f$$ be an $$(\frak{M}\otimes\frak{N})$$-measurable function on $$X\times Y$$. Then if $$f\geq0$$ and $$1\leq p<\infty,$$ we have $$\left[\int\left(\int f(x,y)d\nu(y)\right)^p d\mu(x)\right]^{1/p}\leq\int\left[\int f(x,y)^p d\mu(x)\right]^{1/p}d\nu(y).$$