# Functional from $(0,+\infty)$ to $L^p(\mathbb R)$

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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In the questions a) and b), is it $f$ instead of $f^r$ ? And what did you try ? –  Davide Giraudo Aug 29 '11 at 15:17
I am somewhat tempted to replace $]p,q[$ with $(p,q)$... –  Ｊ. Ｍ. Aug 29 '11 at 15:28
i've edited thanks davide. I've tried to calculate the $L^p$ norm and find a lipschitz constant. I've plugged the integral into mathematica but the expression is in terms of the gamma function so i was looking for something more manageable –  uforoboa Aug 29 '11 at 16:08
Did you compute $\lVert f^{r+h}-f^r\lVert_{L^p}^p$ for $h>0$? Maybe you don't need to compute the integrals, but only to use some inequalities. –  Davide Giraudo Aug 29 '11 at 17:19
I am trying to understand how $f^r$ is defined: maybe you wanted to say that for every $r \in (0, \infty)$, $f^r$ is in $L^p(\mathbb{R})$. That's different from saying that for every $r$ we have $f^r : (0,\infty) \to L^p(\mathbb{R})$. –  Angelo Lucia Aug 29 '11 at 17:55