Need some explanation to this Problem. I was reading a theorem which says : 

If  $ f \in L^p(\mathbb{R}^n) $ and $\Phi\in C^\infty(\mathbb{R}^n)$ then $f_\epsilon= \Phi_\epsilon \star f$ tends to $f$ as $\epsilon \to 0$. 

(here, page number 2). But I am not able to understand the steps. Can anyone explain it to me? I understood the procedure but I am not able to understand the steps. 
Thanks.
 A: Using the definition of the constant $a$ and of the convolution product, we have that
\begin{align*}
(f \ast \varphi_t)(x) - af(x) & = (f \ast \varphi_t)(x) - f(x)\int_{\mathbb{R}^n} \varphi(y) ~dy \\
 & = \int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy - \int_{\mathbb{R}^n} f(x)\varphi(y) ~dy.
\end{align*}
Now in the first integral, make the change of variables $y \mapsto tz$. Then $dy \mapsto t^n ~dz$ (because $tz = (tz_1, \dots, tz_n)$ and $dz$ is really shorthand for $dz_1 \cdots dz_n$ as we are working in $\mathbb{R}^n$) and
$$\varphi_t(tz) = t^{-n} \varphi\left( \frac{tz}{t} \right) = t^{-n}\varphi(z).$$
Hence the first integral becomes
\begin{align*}
\int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy & = \int_{\mathbb{R}^n} f(x - tz)t^{-n}\varphi(z)t^n ~dz \\
 & = \int_{\mathbb{R}^n} f(x - tz)\varphi(z) ~dz.
\end{align*}
So, returning to our original calculation, we have
\begin{align*}
(f \ast \varphi_t)(x) - af(x) & = \int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy - \int_{\mathbb{R}^n} f(x)\varphi(y) ~dy \\
& = \int_{\mathbb{R}^n} f(x - tz)\varphi(z) ~dz - \int_{\mathbb{R}^n} f(x)\varphi(z) ~dz \\
 & = \int_{\mathbb{R}^n} (f(x - tz) - f(x))\varphi(z) ~dz \\
 & = \int_{\mathbb{R}^n} ((\tau_{tz}f)(x) - f(x))\varphi(z) ~dz \\
 & = \int_{\mathbb{R}^n} (\tau_{tz}f - f)(x)\varphi(z) ~dz. \\
\end{align*}
Now $\|\tau_{tz}f - f\|_p \longrightarrow 0$ by "Exercise I.6" in the notes you linked. The rough idea of the proof is to use the density of $C_c(\mathbb{R}^n)$ (compactly supported continuous real-valued functions on $\mathbb{R}^n$) in $L^p(\mathbb{R}^n)$. Given $\varepsilon > 0$, let $g \in C_c(\mathbb{R}^n)$ be such that
$$\|f - g\|_p < \varepsilon.$$
Clearly since $g$ is continuous, for $t$ small enough
$$\|\tau_{tz}g - g\|_p < \varepsilon.$$
Now we have that
\begin{align*}
\|\tau_{tz}f - f\|_p & = \|\tau_{tz}f - \tau_{tz}g + \tau_{tz}g - g + g - f\|_p \\
 & \leq \|\tau_{tz}f - \tau_{tz}g\|_p + \|\tau_{tz}g - g\|_p + \|g - f\|_p \\
 & \leq \|\tau_{tz}\|_{\text{op}}\|f - g\|_p + \|\tau_{tz}g - g\|_p + \|f - g\|_p \\
 & = \|f - g\|_p + \|\tau_{tz}g - g\|_p + \|f - g\|_p \\
 & < 3\varepsilon
\end{align*}
for $t$ small enough. Therefore
$$\|\tau_{tz}f - f\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0.$$
Since
$$\|\tau_{tz}f - f\|_p \leq \|\tau_{tz}f\|_p + \|f\|_p = 2\|f\|_p$$
and
$$\|\tau_{tz}f - f\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0,$$
the hypotheses of the Dominated Convergence Theorem are satisfied and hence
$$\|f \ast \varphi_t - af\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0,$$
or in other words $f \ast \varphi_t \longrightarrow af$ in $L^p(\mathbb{R}^n)$ as $t \longrightarrow 0$.

Addendum: $\tau_{tz}$ is the translation operator
$$\tau_{tz}: L^p(\mathbb{R}^n) \longrightarrow L^p(\mathbb{R}^n),$$
$$(\tau_{tz}f)(x) = f(x - tz).$$
To see why it has norm $1$, note that
\begin{align*}
\|\tau_{tz}\|_{\text{op}} & = \sup_{\|f\|_p = 1} \|\tau_{tz}f\|_p \\
 & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |(\tau_{tz}f)(x)|^p ~dx\right)^{1/p} \\
 & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |f(x - tz)|^p ~dx\right)^{1/p} \\
 & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |f(u)|^p ~du\right)^{1/p} \\
 & = \sup_{\|f\|_p = 1} \|f\|_p \\
 & = 1.
\end{align*}
Above I used the change of variables $x - tz \mapsto u$. So $\|\tau_{tz}\|_{\text{op}} = 1$ is really just a consequence of the translation invariance of the Lebesgue measure.
