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Disclaimer. This post is a partial solution to my problem.

Claim. Suppose $X$ has density and $\mathbb E\,\|X\| \lt \infty$. Then, $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then for every $w \in \mathbb R^n$, the random variable $X' w$ also has density, and hence a continuous CDF. Let $R$ be a large positive number. By the preceding argument function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \ge R-|X'w|$, and so $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t) = \mathbb P(\|X\|_q \ge R - t)\\ &\le \frac{\mathbb E\, \|X\|_q}{R-t} \lesssim \frac{1}{R} \overset{R \to \infty}{\longrightarrow} 0. \end{split} \tag{2} $$

Combining (1) and (2) completes the claim.

Claim. If $X$ has density, then $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then so does $F(X)$, for any continuous function $F:\mathbb R^n \to \mathbb R^m$. In particular, for every $w \in \mathbb R^n$, the random variable $X' w$ has density, and hence a continuous CDF. Let $R$ be a large positive number. By compacity of $S_p^n \times [-R,R]$ and the preceding argument, the function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \gt R-|X'w|$, and so for any $t \ge 0$, one computes $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t)\\ &= \mathbb P(\|X\|_q \ge R - t) \longrightarrow 0 \text{ in the limit }R \to \infty. \end{split} \tag{2} $$

In the last step, we have used the fact that the CDF of $\|X\|_q$ is continuous (because $X$ has density and so $\|X\|_q$ does too, by continuity of the $\ell_q$-norm on $\mathbb R^n$). Combining (1) and (2) completes the claim.

Disclaimer. This post is a partial solution to my problem.

Claim. Suppose $X$ has density and $\mathbb E\,\|X\|_q \lt \infty$, where $q$ is the Hoelder / harmonic conjugate of $p$. Then, $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then for every $w \in \mathbb R^n$, the random variable $X' w$ also has density, and hence a continuous CDF. Let $R$ be a large positive number. By the preceding argument function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \ge R-|X'w|$, and so $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t) = \mathbb P(\|X\|_q \ge R - t)\\ &\le \frac{\mathbb E\, \|X\|_q}{R-t} \lesssim \frac{1}{R} \overset{R \to \infty}{\longrightarrow} 0. \end{split} \tag{2} $$

Combining (1) and (2) completes the claim.

Disclaimer. This post is a partial solution to my problem.

Claim. Suppose $X$ has density and $\mathbb E\,\|X\| \lt \infty$. Then, $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then for every $w \in \mathbb R^n$, the random variable $X' w$ also has density, and hence a continuous CDF. Let $R$ be a large positive number. By the preceding argument function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \ge R-|X'w|$, and so $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t) = \mathbb P(\|X\|_q \ge R - t)\\ &\le \frac{\mathbb E\, \|X\|_q}{R-t} \lesssim \frac{1}{R} \overset{R \to \infty}{\longrightarrow} 0. \end{split} \tag{2} $$

Combining (1) and (2) completes the claim.

Disclaimer. This post is a partial solution to my problem.

Claim. Suppose $X$ has density and $\mathbb E\,\|X\|_q \lt \infty$, where $q$ is the Hoelder / harmonic conjugate of $p$. Then, $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then for every $w \in \mathbb R^n$, the random variable $X' w$ also has density, and hence a continuous CDF. Let $R$ be a large positive number. By the preceding argument function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \ge R-|X'w|$, and so $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \sup_{w \in S_p^n,\,|u| \gt R}\mathbb P(|u| - |X'w| \le t)\\ & \le \sup_{w \in S_p^n,\,|u| \gt R}\mathbb P(R - |X'w| \le t)\\ & \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t)\\ &= \mathbb P(\|X\|_q \ge R - t)\\ &\le \frac{\mathbb E\, \|X\|_q}{R-t} \lesssim \frac{1}{R} \overset{R \to \infty}{\longrightarrow} 0. \end{split} \tag{2} $$

Combining (1) and (2) completes the claim.

Disclaimer. This post is a partial solution to my problem.

Claim. Suppose $X$ has density and $\mathbb E\,\|X\|_q \lt \infty$, where $q$ is the Hoelder / harmonic conjugate of $p$. Then, $L(S_p^n,t) \longrightarrow 0$ in the limit $t \to 0^+$.

Indeed, if $X$ has density, then for every $w \in \mathbb R^n$, the random variable $X' w$ also has density, and hence a continuous CDF. Let $R$ be a large positive number. By the preceding argument function $t \mapsto \underset{w \in S_p^n,\,|u| \le R}{\sup}\mathbb P(|X'w-u| \le t)$ is continuous, and so $$ \tag{1} \lim_{t \to 0^+}\sup_{w \in S_p^n}\sup_{|u| \le R}\mathbb P(|X'w-u| \le t) = \sup_{w \in S_p^n,\,|u| \le R}\lim_{t \to 0^+}\mathbb P(|X'w-u| \le t) = 0. $$

On the other hand, if $|u| \gt R$, then $|X'w-u| \ge ||X'w| - |u|| \ge |u| - |X'w| \ge R-|X'w|$, and so $$ \begin{split} \sup_{w \in S_p^n}\sup_{|u| \gt R} \mathbb P(|X'w-u| \le t) &\le \mathbb P(R - |X'w| \le t) \le \sup_{w \in S_p^n} \mathbb P(|X'w| \ge R - t)\\ & \le \mathbb P(\sup_{w \in S_p^n} |X'w| \ge R - t) = \mathbb P(\|X\|_q \ge R - t)\\ &\le \frac{\mathbb E\, \|X\|_q}{R-t} \lesssim \frac{1}{R} \overset{R \to \infty}{\longrightarrow} 0. \end{split} \tag{2} $$

Combining (1) and (2) completes the claim.