Limiting property of spherical mean value operator

Question

Let $M_t$ denotes the spherical mean value operator on $R^n$. Let $\{f_n\}_{n \in N}$ be a sequence of real valued smooth functions on $R^n$ converges to $f$ uniformly over compact set. Further assume that $M_tf_n$ converges to $f_n$ uniformly on compact sets as $t \rightarrow \infty$. Then show that
$M_tf$ converges to $f$ uniformly on compact sets as $t \rightarrow \infty$.

Answer 1

DISCLAIMER: This solution has a flaw as of now. Im gonna leave it up since the mistake has been pointed out. Ill re-visit when I find the time
!!

Lets denote the your sphere by $\mathbb{S}^{n-1}_t$, where t is the radius, and note that it is compact. Now we have uniform convergence on compact sets and we know that we can take the limit of we can exchange limit and integration now for all $t >0 $:
$$ lim_{n \to \infty}M_tf_n=M_t(lim_{n \to \infty}f_n) $$ Now we can estimate for a fixed $t > 0$ using the triangle-inequality:
$$ 0 \leq |M_tf-f|\leq|M_tf-M_tf_n|+|M_tf_n-f_n|+|f_n-f| $$ Using that we can pull the limit under the integral, we see that: $$ 0 \leq |M_tf-M_tf_n| \leq \sup |(f-f_n)|Vol(\mathbb{S}^{n-1}_t)\frac{1}{Vol(\mathbb{S}^{n-1}_t)}=\sup |(f-f_n)| $$ and it is indpendant of t. We can now procede to estimate, by taking $t \to \infty $ $$ 0 \leq |M_tf-f| \leq 2|f-f_n|+|M_tf_n-f_n| \to2|f-f_n| $$ Taking now the limit with respect to $n \to \infty$ gives us the result: $$ 0 \leq |M_tf-f|\leq2|f_n-f| \to 0 $$

Answer 2

ur answer has mistake i.e. $|M_t f-M_tf_n|\leq sup |f-f_n|$, here supremum is taken over $S_t^{n-1}$, hence right side depends on $t$. Therefore ur argument doesn't work.