We have this theorem
For any $1\le p<\infty$ and $f\in L^p(\mathbb{R}^k)$, then $\|f*\phi_\delta - f\|_p\to 0$ as $\delta\to0$, where $\phi$ is any nonnegative measurable function on $\mathbb{R}^k$ with total integral one.
and the proof:
We need to prove that $$\lim_{\delta\to0}\|f*\phi_\delta-f\|_p=0.$$
we have $$|(f*\phi_\delta)(x)-f(x)|\le \int_{\mathbb{R}^k}|f(x-y)-f(x)|\phi_\delta(y)\,dm(y).$$ Since $\phi_\delta$ is a positive function with integral one, we may apply Jensen's Inequality for the convex function $t\mapsto t^p$ to find \begin{eqnarray*} |(f*\phi_\delta)(x)-f(x)|^p &\le& \left(\int_{\mathbb{R}^k} |f(x-y)-f(x)|\phi_\delta (y)\,dm(y)\right)^p \\ &\le& \int_{\mathbb{R}^k} |f(x-y)-f(x)|^p\phi_\delta (y)\,dm(y). \end{eqnarray*} Now we may integrate this inequality over $\mathbb{R}^k$ in $x$ and use Fubini's Theorem to find \begin{eqnarray*} \|f*\phi_\delta-f\|_p^p &\le&\int_{\mathbb{R}^k}\int_{\mathbb{R}^k}|f(x-y)-f(x)|^p \phi_\delta(y)\,dm(y)\,dm(x) \\ &=& \int_{\mathbb{R}^k} \int_{\mathbb{R}^k}|f(x-y)-f(x)|^p\,dm(x)\, \phi_\delta(y)\,dm(y) \\ &=&\int_{\mathbb{R}^k} \|f^y-f\|_p^p\,\phi_\delta(y)\,dm(y). \end{eqnarray*} Let $g(y)=\|f^y-f\|_p^p$. $g$ is a continuous function, and it is clear that $g(0)=0$. Moreover, $g$ is bounded since $$g(y)=\|f^y-f\|_p^p \le(\|f^y\|_p + \|f\|_p)^p = (2\|f\|_p)^p.$$ We continue computing to find \begin{eqnarray*} \|f*\phi_\delta-f\|_p^p &\le& \int_{\mathbb{R}^k}g(y)\phi_\delta(y)\,dm(y) \\ &=& \int_{\mathbb{R}^k}g(y)\delta^{-k}\phi(y\delta^{-1})\,dm(y), \\ &=& \int_{\mathbb{R}^k}g(\delta s)\phi(s)\,dm(s) \end{eqnarray*} for the change of variables $s=y\delta^{-1}$. As $\delta\to0$, $g(\delta s)\phi(s)\to g(0)\phi(s)=0$ pointwise on $\mathbb{R}^k$. Moreover, the integrand $g(\delta s)\phi(s)\le \|g\|_\infty \phi(s)$ for all $\delta$. Since $g$ is bounded and $\phi$ is integrable, the Dominated Convergence Theorem applies to show that $$\lim_{\delta\to0}\int_{\mathbb{R}^k} g(\delta s)\phi(s)\,dm(s)=0.$$ This implies $\|f*\phi_\delta-f\|_p\to0$ as $\delta\to0$.
My problem is that I do not understand how it applies the Dominated Convergence Theorem at the end of the proof. Can anybody help me?
What we need is the following: if $(f_\delta)_{\delta\in (0,1]}$ is a family of function indexed by a real number $\delta$ such that the function $t\mapsto \sup_{\delta\in(0,1]}\left|f_\delta(t)\right|$ is integrable and for each $t$, $\lim_{\delta\to 0}f_\delta(t)=0$, then $\lim_{\delta\to 0}\int_{\mathbb R}f_\delta(t)\mathrm d\lambda(t)=0$.
To prove this, assume that $\int_{\mathbb R}f_\delta(t)\mathrm d\lambda(t)$ does not converge to $0$ as $\delta$ goes to $0$. This means that there exists some positive $\varepsilon$ and a sequence $\left(\delta_n\right)_{n\geqslant 1}$ which converges to $0$ such that for each $n$, $$\tag{*}\left|\int_{\mathbb R}f_{\delta_n}(t)\mathrm d\lambda(t)\right|\gt \varepsilon.$$ Now, the sequence $\left(f_{\delta_n}\right)_{n\geqslant 1}$ satisfies the assumptions of the classical dominated convergence theorem, hence (*) entails a contradiction.