Is my proof about the behavior of Convolutions as $||x|| \rightarrow \infty$ correct?

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I've written a proof to the following statement and would appreciate if somebody could look over it and give some feedback. My apologies if this is the wrong place for this.

Statement

Let $f \in L^p(\mathbb{R}^n), g \in L^q(\mathbb{R}^n)$ such that $\frac{1}{p} + \frac{1}{q} = 1$, show that $(f * g)(x) \rightarrow 0$ as $||x||_2 \rightarrow \infty$.

My Proof

$C := \max \{||f||_p, ||g||_q\}$. Let $f \in L^p (\mathbb{R}^n)$ and consider $f_k := \left|\left.f\right|_{\mathbb{R}^n \backslash B_{k}(0)}\right|^p \overset{k \rightarrow \infty}{\longrightarrow} 0$ pointwise. Obviously this function is dominated by $|f|^p$ and therefore by the dominated convergence theorem $f_k \rightarrow 0$ in $L^p(R^n)$.

Because of this we can find a sequence $(R_n)_{n} \subseteq \mathbb{R}_{> 0}$ such that $R_{n+1} > R_{n} > 0$ and \begin{align*} \max \{||\left. f\right|_{B_n^c}||_p, ||\left. g\right|_{B_n^c}||_q\} < \frac{1}{2n} \tag{*} \end{align*} where $B_n := B_{R_n}(0), B_n^c := \mathbb{R}^n \backslash B_n$.

Now let $(x_n) \subseteq \mathbb{R}^n$ be some sequence such that $||x_n||_2 \in (2 R_n, 2 R_{n+1})$. $T_ng(y) := g(x_n - y)$. \begin{align*} |f * g(x_n)| &= \left| \int_{\mathbb{R}^n} f(y) g(x_n - y) d\lambda^n(y) \right| \\ &\leq \int_{\mathbb{R}^n} |f(y) T_ng(y)| d\lambda^n(y) \\ &= \int_{B_n} |f(y) T_ng(y)| d\lambda^n(y) + \int_{B_n^c} |f(y) T_ng(y)| d\lambda^n(y) \\ &= ||f \cdot T_ng||_{L^1(B_n)} + ||f \cdot T_ng||_{L^1(B_n^c)} \\ &\overset{\text{Hölder}}{\leq} ||f||_{L^p(B_n)} \cdot ||T_ng||_{L^q(B_n)} + ||f||_{L^p(B_n^c)} \cdot ||T_g||_{L^q(B_n^c)} \\ &\overset{(*)}{\leq} ||f||_{L^p(B_n)} \cdot \frac{1}{2n} + ||T_g||_{L^q(B_n^c)} \cdot \frac{1}{2n} \leq \frac{C}{n} \longrightarrow 0. \end{align*}