We use the notation $\mathbb{K}$ to mean either the real number field $\mathbb{R}$ or the complex number field $\mathbb{C}$. Let $X$ be a locally compact group and let $\lambda$ denote the unique Haar Measure on the Borel sigma algebra of $X$. Recall that an approximate identity is a family $(u_\epsilon)_{\epsilon > 0} \subset L^1(X, \mathbb{K})$ satisfying
a) $\exists c>0$ such that $\|u_\epsilon\|_1 = \int_X |u_\epsilon| d\lambda < c$
b) $\forall \epsilon >0 \colon \int_X u_\epsilon = 1$
c) for any neighborhood $U$ of the neutral element $e \in X$ $$\int_{X\backslash U} |u_\epsilon| d\lambda \to 0$$ for $\epsilon \to 0$.
$\textbf{Lemma}$: Let $1 \leq p < \infty$ and suppose $f \in L^p(X, \mathbb{K})$, then for all $\delta>0$ there exists an open neighborhood $U$ of $e$ so that $$\int_X |f(x)-f(y^{-1}x)|^p d\lambda(x) \leq \delta$$ for all $y \in U$.
$\textbf{Theorem}$: If $(u_\epsilon)_{\epsilon >0}$ is an approximate identity, $1 \leq p < \infty$ and $f \in L^p(X, \mathbb{K})$, then $$u_\epsilon \star f \to_{\|.\|_p} f$$ where $\to_{\|.\|_p}$ denotes convergence in $L^p(X, \mathbb{K})$ wrt. the $L^p$-norm and $\star$ denotes the convolution, that is $g \star h(z) = \int_X g(x)h(x^{-1}z)d\lambda(x)$.
$\textbf{Proof:}$ For $p=1$ note, that \begin{eqnarray} \|f-u_\epsilon \star f\|_1 = \int |f-u_\epsilon \star f| d\lambda(x) = \int \bigg|f(x)\int u_\epsilon(y) d\lambda(y)- \int u_\epsilon(y)f(y^{-1}x) d\lambda(y) \bigg| d\lambda(x) \\ \leq \int \int |f(x)-f(y^{-1}x)| d\lambda(x) |u_\epsilon(y)| d\lambda(y) \end{eqnarray} Splitting $X = (X \backslash U) \cup U$ by means of the lemma for some $\delta>0$ leads to
\begin{eqnarray} \|f-u_\epsilon \star f\|_1 \leq \int_U \int |f(x)-f(y^{-1}x)| d\lambda(x) |u_\epsilon(y)| d\lambda(y) + \int_{X \backslash U} \int |f(x)-f(y^{-1}x)| d\lambda(x) |u_\epsilon(y)| d\lambda(y) \\ \leq \delta \int |u_\epsilon| d\lambda(y) + 2 \|f\|_1 \int_{X\backslash U} |u_\epsilon| d\lambda(y) \leq \delta c + 2 \|f\|_1 \int_{X \backslash U} |u_\epsilon| d\lambda(y) \end{eqnarray} which can be made arbitrarily small by definition of an approximate identity. Now the author claims, that the case of $1 < p < \infty$ is analogous upon noting, that $\|\|h\|_1\|_p \leq \|\|h\|_p\|_1$ for a function $h$ in two variables.
I am somehow stuck at the case for $1 < p < \infty$ though. Since if we use $\|\|f\|_1\|_p \leq \|\|f\|_p\|_1$ then we can estimate:
$\begin{eqnarray} \|f-u_\epsilon \star f\|_p = \big(\int |f-u_\epsilon \star f|^p d\lambda(x) \big)^{1/p} = \big(\int \bigg|\int \underbrace{(f(x)-f(y^{-1}x))u_\epsilon(y)}_{ = h(x,y)} d\lambda(y)\bigg|^p d\lambda(x) \big)^{1/p} \leq \big(\int \bigg|\int |h(x,y)| d\lambda(y)\bigg|^p d\lambda(x) \big)^{1/p} = \|\|h\|_1\|_p \leq \|\|h\|_p\|_1 = \int \big(\int |f(x)-f(y^{-1}x)|^p|u_\epsilon(y)|^p d\lambda(x)\big)^{1/p} d\lambda(y) \end{eqnarray}$
Now I guess we would be partitioning $X = (X\backslash U) \cup U$ as given in the lemma again. However my problem now is, that $\int |u_\epsilon|^p d\lambda(y)$ need not be finite in general (or does it?) and in addition we do not know if $\int_{X \backslash U} |u_\epsilon|^p d\lambda(y) \to 0$ as $\epsilon \to 0$. So how can we conclude our statement holds in the case of $1 < p < \infty$, since this doesn't seem to be analogous at all?
Does someone have a reference for why $\|\|h\|_1\|_p \leq \|\|h\|_p\|_1$ even holds true?
$\textbf{Edit:}$ In the appendix of the lecture notes the aforementioned inequality is stated as follows:
For $1 \leq p <\infty$ and $h \colon X \times Y \to \mathbb{K}$ we have $$\|\|h\|_1\|_p \leq \|\|h\|_p\|_1$$ where $\|.\|_1$ and $\|.\|_p$ are with respect to $X$ and $Y$, respectively.
Let $1 \leq p < \infty$ and let $f \in L^p(X, \mathbb{K})$, then it can be seen, either by direct verification or a duality argument, that $$\bigg(\int |f|^p d\lambda \bigg)^{1/p} = \max\limits_{\substack{g \in L^q(X, \mathbb{K}) \\ \|g\|_q = 1}} \bigg| \int fg d\lambda \bigg|$$ where $q$ is such that $1/p+1/q = 1$ and $\max$ indicates, that the supremum is attained. Having this in our toolbox we can state and prove the inequality
$\textbf{Minkowski's integral inequality:}$ Let $K(x,y)$ be a positive, measurable kernel on some product space $(\Omega_1 \times \Omega_2, \mu \times \nu)$. If $p \geq 1$, then $$\bigg(\int_{\Omega_1} \bigg(\int_{\Omega_2} K(x,y) d\nu \bigg)^p d\mu \bigg)^{1/p} \leq \int_{\Omega_2} \bigg(\int_{\Omega_1} K(x,y)^p d\mu \bigg)^{1/p} d\nu$$ $\mathbf{Proof:}$ If $p = 1$ the above inequality turns into an equality by Fubini. So let $p > 1$ and note that there exists $g \in L^q(X, \mathbb{K})$ with $\|g\|_q = 1$ such that $$\bigg(\int_{\Omega_1} \bigg(\int_{\Omega_2} K(x,y) d\nu \bigg)^p d\mu \bigg)^{1/p} = \int_{\Omega_1}g(x) \int_{\Omega_2} K(x,y) d\nu(y) d\mu(x)$$ Utilizing Fubini we note that
$\begin{eqnarray} \int_{\Omega_1}g(x) \int_{\Omega_2} K(x,y) d\nu(y) d\mu(x) = \int_{\Omega_2} \int_{\Omega_1}g(x) K(x,y) d\mu(x) d\nu(y)\\ \leq \int_{\Omega_2} \sup\limits_{\|h\|_q = 1} \bigg| \int_{\Omega_1} h(x) K(x,y) d\mu(x) \bigg| d\nu(y) = \int_{\Omega_2} \bigg(\int_{\Omega_1} K(x,y)^p d\mu \bigg)^{1/p} d\nu \end{eqnarray}$
as desired.
Now we get back to the main question and estimate
$\begin{eqnarray} \|f-u_\epsilon \star f\|_p = \bigg(\int |f(x)-u_\epsilon \star f(x)|^p d\lambda(x) \bigg)^{1/p} = \bigg(\int \bigg|\int \big(f(x)-f(y^{-1}x)\big)u_\epsilon(y) d\lambda(y)\bigg|^p d\lambda(x) \bigg)^{1/p} \leq \int \bigg(\int |f(x)-f(y^{-1}x)|^p |u_\epsilon(y)|^p d\lambda(x) \bigg)^{1/p} d\lambda(y) = \int \bigg(\int |f(x)-f(y^{-1}x)|^p d\lambda(x) \bigg)^{1/p} |u_\epsilon(y)| d\lambda(y) \end{eqnarray}$ which shows that as in the case of $p = 1$ we can split $X$ into $X = (X \backslash U) \cup U$ to verify, that $\|f - u_\epsilon \star f\|_p$ gets arbitrarily small for $\epsilon \to 0$. The real struggle with this question was, that i didn't notice the variables switching in the integral inequality, which is essential for the estimations.