Average of a $L^p$ function in the ball $B_r(x) \subset \mathbb{R}^n$ using convolution results
Let $1 \le p < \infty$. For all $r >0$ let $T_r : L^p(\mathbb{R}^n) \rightarrow L^p(\mathbb{R}^n)$ the operator defined as follows
$T_r f(x): = \frac{1}{|B_r(x)|} \int_{B_r(x)} f(y) dy$, for $x \in \mathbb{R}^n, f \in L^p(\mathbb{R}^n)$
Prove that:
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$T_r$ is well-defined, that is $T_r f \in L^p(\mathbb{R}^n)$ if $f \in L^p(\mathbb{R}^n)$.
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$T_r \in \mathcal{L}( L^p(\mathbb{R}^n))$ and calculate $||T_r||_{\mathcal{L}( L^p(\mathbb{R}^n))}$.
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For all $f \in L^p(\mathbb{R}^n)$, $T_r f \in C^0(\mathbb{R}^n)$ and $T_r f(x) \rightarrow 0$ as $|x| \rightarrow \infty$.
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$T_rf \rightarrow f$ for $r \rightarrow 0^+$, for all $f \in L^p(\mathbb{R}^n)$.
The suggestion was to write $T_r f = \phi_r * f$ for a suitable $\phi_r$. After some manipulations, I've found that $T_r f(x): = \int_{\mathbb{R}^n} f(y) \frac{\chi_{B_r(0)}(x-y)}{|B_r(0)|} dy = \phi_r * f$, where $\phi_r (t) := \frac{\chi_{B_r(0)}(t)}{|B_r(0)|} \in L^1(\mathbb{R}^n)$. Then, $T_rf \in L^p(\mathbb{R}^n)$ and $||T_r f||_{L^p} \le ||\phi_r||_{L^1} ||f||_{L^p} = ||f||_{L^p}$ (since it is immediate to prove that $||\phi_r||_{L^1} = 1$). $T_r$ is linear (obvious) and continuous since it holds $||T_r f||_{L^p} \le ||f||_{L^p}$, for all $f \in L^p(\mathbb{R}^n)$. From this reasoning, it follows also that $||T||_{\mathcal{L}( L^p(\mathbb{R}^n))} \le 1$. I'd like to prove that the norm is actually $1$ but I am struggling to find a function which does the job.
As for what concerns point 3. my idea was to proceed using the following property of the convolution. I know that $\phi_r \in L^1(\mathbb{R}^n) \Rightarrow \phi_r \in L^1_{loc}(\mathbb{R}^n)$ and that continuous functions with compact support in $\mathbb{R}^n$ are dense in $L^p(\mathbb{R}^n)$ for $1 \le p < \infty$. Thus, I can approximate $f \in L^p(\mathbb{R}^n)$ with a suitable sequence $(f_n)_n$ of functions in $C^0_c(\mathbb{R}^n)$ and consider $\phi_r * f_n \in C^0(\mathbb{R}^n)$. However, I don't know whether I can conclude from here or if I need further assumptions. Once I have shown that $T_r f \in C^0(\mathbb{R}^n)$, I think that being $T_r f \in L^1(\mathbb{R}^n)$ and since the limit exists by continuity, I can conclude that $\lim_{|x| \rightarrow \infty} T_r f(x) = 0$.
I know that 5. follows from a sort of generalization of the fundamental theorem of calculus, but I have few ideas on how to generalize it formally in dimensions higher than 1.
Any hint or suggestion on how to proceed would be greatly appreciated.
Partial proof.
Let $1<p<\infty$ and $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then $$ |(T_rf)(x)|=\frac{1}{|B_r(x)|}\left|\int_{B_r(x)}f(y)\,dy\,\right| \le \frac{1}{|B_r(x)|}\left|\int_{B_r(x)}1\cdot |f(y)|\,dy\,\right| \\ \le \frac{1}{|B_r(x)|}\left(\int_{B_r(x)}1^q\right)^{1/q}\cdot \left(\int_{B_r(x)}|f(y)|^p\,dy\,\right)^{1/p}\\ =w_n^{1/q}r^\frac{n}{q}\frac{1}{\omega_nr^n}\left(\int_{B_r(x)}|f(y)|^p\,dy\,\right)^{1/p} =\frac{1}{\omega_n^{1/p}r^{n/p}} \left(\int_{B_r(x)}|f(y)|^p\,dy\,\right)^{1/p} $$ Hence $$ |(T_rf)(x)|^p\le \frac{1}{\omega_nr^n}\int_{B_r(x)}|f(y)|^p\,dy =\frac{1}{\omega_nr^n}\int_{B_r(0)}|f(x+w)|^p\,dw $$ and thus $$ \int_{\mathbb R^n}|(T_rf)(x)|^p\,dx\le \frac{1}{\omega_nr^n}\int_{\mathbb R^n}\int_{B_r(0)}|f(x+w)|^p\,dw\,dx= \frac{1}{\omega_nr^n}\int_{B_r(0)}\int_{\mathbb R^n}|f(x+w)|^p\,dw\,dx \\ =\frac{1}{\omega_nr^n}\omega_nr^n\|f\|^p=\|f\|^p, $$ and finally $$ \|T_rf\|_p\le \|f\|_p. $$ The case $p=1$ is simpler.