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For every $r\in(0,+\infty)$, we define $f^r:\mathbb R\to\mathbb R$ to be

$f^r(x)=\begin{cases}\sqrt{r^2-x^2}&\text{if } |x|\leq r\\ 0 & \text{otherwise} \end{cases}$

a)Find all $p\geq 1$ such that the map $r\to f^r$ from $(0,+\infty)$ to $L^p(\mathbb R)$ is continuous

b)Find all $p\geq 1$ such that the map $r\to f^r$ from $(0,+\infty)$ to $L^p(\mathbb R)$ is differentiable

-Mario-

Edit

sorry everybody... i miscopied the text... i edited.. really i apologize..

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    I would like to put my own two cents regarding part a). Like Davide Giraudo suggested one calculates $\|f^{r+h}-f^r\|_{L^p}^p=2\int_{-r-h}^{-r}(\sqrt{(r+h)^2-x^2})^p\mathrm d x+\int_{-r}^{r}\left(\frac{2rh+h^2}{\sqrt{(r+h)^2-x^2}+\sqrt{r^2-x^2}}\right)^p\mathrm d x\leq 2h(r+h)^p+2rh^p$. Analogously $\|f^r-f^{r-h}\|_{L^p}^p\leq 2hr^p+2(r-h)h^p$. Hence continuity is ensured for any $p\geq 1$. Am I right?2011-08-30

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