Given the function
$$ u(x,t)=\frac{1}{2 \Vert x \Vert}\int_{\vert \Vert x\Vert -t \vert}^{\Vert x\Vert +t}{s h(s)ds} $$
where $h \in C^2(\mathbb R)$, $h \in L^2(\mathbb{R}^3)$ and $x \in \mathbb{R}^3$.
We define the maximal function $$(Mf)(t) = \sup_{r>0}{\frac{1}{2r}\int_{-r}^r}{\vert f(t-r) \vert}dr$$
Given the property $\Vert M(f)\Vert_{L^2(\mathbb{R}^n)} \le \Vert f\Vert_{L^2(\mathbb{R}^n)}$ for every $f \in L^2(\mathbb{R}^n)$ show that $$ \int_{\mathbb{R}}{\Vert u(t,\cdot)\Vert_{L^{\infty}(\mathbb{R}^n)}^2} \le C \Vert h \Vert_{L^2(\mathbb{R}^n)}^2 $$
This almost looks like a direct application of the inequality for maximal functions but the integrand of $u$ is not quite right. How can I apply it here ?
Would appreciate any help/hints
Several things should be clarified. It seems that there are many typos on your questions. If I change the definition of $u$, then I can show that the desired assertion(modified one) holds. Since it is too long to explain in the comment, I write it here. I'm not sure that why $s$ came in the definition of $u$. So I omit it.
Let $h\in L^2 (\mathbb{R})$. For $x\in \mathbb{R}^n$, we define $$ u(x,t) = \begin{cases} \frac{1}{2\Vert x \Vert} \int_{t-\Vert{x}\Vert}^{t+\Vert{x}\Vert} h(s) ds&\text{if } x\neq 0,\\ 0 & \text{if } x=0. \end{cases} $$ Our definition is well-defined since $h$ is locally integrable. Then by definition of maximal function, we have $$ |u(x,t)| \leq M(h)(t)\quad \text{a.e. on } x.$$ Hence it follows from $L^2$-boundedness of Hardy-Littlewood maximal function that $$ \int_{-\infty}^\infty \Vert u(\cdot,t)\Vert_{L^\infty(\mathbb{R}^n)}^2 dt\leq \int_{-\infty}^\infty |M(h)(t)|^2 dt \leq C \int_{-\infty}^\infty |h(t)|^2 dt=C\Vert h \Vert_{L^2(\mathbb{R})}^2. $$
If you correct typos on your original question, let me know.