Question on Stein and Shakarchi's proof of Proposition 2.5 ($f(x-h) \rightarrow f$ in $L^1$ as $h \rightarrow 0$)
My question is regarding the proof of Proposition 2.5 (p.74) in Stein & Shakarchi's Real Analysis:
Proposition 2.5: Suppose $f \in L^1(\mathbb{R}^d)$. Then $$\|f_h - f \|_{L^1} \rightarrow 0 \quad \textit{as } h \rightarrow 0.$$
The proof is a simple consequence of the approximation of integrable functions by continuous functions of compact support as given in Theorem 2.4. In fact for any $\epsilon > 0$, we can find a [continuous and compactly supported] function $g$ so that $\|f - g\| < \epsilon.$ Now $$f_h - f = (g_h - g) + (f_h - g_h) - (f-g).$$
However, $\|f_h - g_h\| = \|f-g\| < \epsilon$, while since $g$ is continuous and has compact support we have that clearly $$\|g_h - g \| = \int_{\mathbb{R}^d} |g(x-h) - g(x)|\,dx \rightarrow 0 \text{ as } h \rightarrow 0.$$ So if $|h| < \delta$, where $\delta$ is sufficiently small, then $\|g_h - g\| < \epsilon$, and as a result $\|f_h-f\| < 3\epsilon$, whenever $|h| < \delta$.
My question: I'm unsure about this step: "We have that clearly
$$\|g_h - g \| = \int_{\mathbb{R}^d} |g(x-h) - g(x)|\,dx \rightarrow 0 \text{ as } h \rightarrow 0$$
..."
Why must this be true?
My thought was to let $$M(h) := \max\limits_{x \in \text{supp}(g)} |g_h(x) - g(x)|$$ and then argue that
- $M(h)$ is defined for all $h$ because $\text{supp}(g)$ is compact, and then
- $M(h) \rightarrow 0$ as $h \rightarrow 0$ (since we clearly have $g_h \rightarrow g$ pointwise).
Is this the right idea? It feels like we are using uniform convergence, but in this case $\{g_h\}_{h \in \mathbb{R}}$ is not a sequence of functions...is it accurate to say that $g_h \rightarrow g$ uniformly as $h \rightarrow 0$? Or is there a different terminology for this type of convergence (aside from pointwise)?
Regarding the notation: $g_h(x) := g(x-h)$ (the translation of $g$ by $h$), and the authors use $\| \cdot \|$ (without the subscript) to denote the $L^1$ norm.
Continuous and compactly supported functions $g_h$ converge uniformly to $g$ as $h\to 0$. Furthermore, as $g$ is compactly supported, the support of $g_h - g$ stays uniformly bounded for $h$ in a bounded set. From this we see that $g_h$ also converges to $g$ in $L^1$ norm.
For the relation between pointwise compactly-supported continuous convergence and uniform convergence, see this link. For the relation between uniform convergence and the convergence in $L^1$-norm see this link.