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Assume $f$ and $g$ are continuous and related to each other as $$ f(x) = \int _{0}^{x-1} \Big ( (x- y)^2 - 1\Big )^{3/2}g(y) \, dy, \qquad x>1. $$

If we happen to know that $f$ is real analytic at some point $x_0$ can we deduce that there is a point $\phi (x_0)$ where $g$ is real analytic? (I suspect $\phi (x_0)$ should be one of the points $x_0 \pm 1$.)

I'm also very interested in incomplete answers such as possible ways to prove this.

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Since $(x-y)^2-1=0$ for $y=x-1$, it follows that $f'''(x)$ does not exist. How do you then expect $f$ to be analytic? – Mercy King Dec 13 '12 at 13:08
up vote 1 down vote accepted

The function $f$ is not analytic. Indeed, setting \begin{eqnarray} \eta(u_1,u_2)&=&\Big[(u_1-u_2)^2-1\Big]^{3/2},\\ \Psi(u_1,u_2)&=&\int_0^{u_2}\eta(u_1,t)g(t)\ dt, \end{eqnarray} we have $$ f(x)=\Psi(x,x-1). $$ It follows that \begin{eqnarray} f'(x)&=&\partial_1\Psi(x,x-1)+\partial_2\Psi(x,x-1)=\int_0^{x-1}\partial_1\eta(x,t)g(t)\ dt+\eta(x,x-1)g(x-1)\\ &=&\int_0^{x-1}\partial_1\eta(x,t)g(t)\ dt=3\int_0^{x-1}(x-t)\sqrt{(x-t)^2-1}\ g(t)\ dt;\\ f^{(2)}(x)&=&\int_0^{x-1}\partial_1^2\eta(x,t)g(t)\ dt+\partial_1\eta(x,x-1)g(x-1)\\ &=&\int_0^{x-1}\partial_1^2\eta(x,t)g(t)\ dt=3\int_0^{x-1}\frac{2(u_1-t)^2-1}{\sqrt{(u_1-t)^2-1}}g(t)\ dt. \end{eqnarray} Since $$ \lim_{t \to x-1}|\partial_1^2\eta(x,t)g(t)|=\infty $$ it is clear that $f^{(3)}(x)$ does not exist. Thus $f$ is not analytic.

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Thank you very much for your input Mercy. I'm a bit puzzled because the same argument works also for $f(x)=\int _{\mathbb{R}}(x-y)^{3/2}_+ g(y) dy$. This has been dealt with elsewhere and they Fourier transformed and inverted the convolution to conclude that $f$ is real analytic near a point $x_0$ iff $g$ is real analytic near $x_0$. – flavio Dec 14 '12 at 9:23
I this example the convolution $\displaystyle f(x)=\int_\mathbb{R}(x-y)_+^{3/2}g(y)dy$ is NOT well defined for any continuous function $g$. In fact you can prove that, if $\lim_{x \to -\infty}g(x)>0$ then $f(x)=\infty$ for every $x$, and if $\lim_{x \to -\infty}g(x)<0$, then $f(x)=-\infty$ for every $x$. Perhaps there is some information you are missing! – Mercy King Dec 14 '12 at 10:26
Yes, sorry, $g$ also has compact support, which is also true for $g$ in my original post. – flavio Dec 14 '12 at 10:51
Unfortunately even if you require $g$ to have compact support in your original post it doesn't make any difference because the problem lies with the fact the at the function the $t \mapsto (t^2-1)^{3/2}$ is NOT twice differentiable at $t=1$, and therefore the function $(x,y) \mapsto [(x-y)^2-1]^{3/2}$ has no partial derivative of order 2 on the line $y=x-1$. – Mercy King Dec 14 '12 at 11:31
I agree, but surely the same holds for $(x-y)^{3/2}$. – flavio Dec 14 '12 at 11:40

Here is a start. the integral equation can be written as

$$ f(x+1) = \int _{0}^{x} \Big ( (x+1- t)^2 - 1\Big )^{3/2} g(t) \, dt$$

$$ = \int _{0}^{x} \Big( ((x-t)+1)^2 - 1\Big )^{3/2}g(t) \, dt, $$

$$\implies f(x+1) = \int _{0}^{x} \Big( (x-t)^2+2(x-t) \Big )^{3/2}g(t) \, dt, $$

Now, the right hand side is a convolution of two functions. You can try the laplace transform technique to solve the integral equations. Also, the above integral equation is known as Volterra integral equation( just let $f(x+1)=h(x)$ ) of the first kind. Solutions of some Volterra equations of the first kind can be found here.

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After the change of variable what you get is $\displaystyle f(x)=\int_1^x(t^2-1)^{3/2}g(x-t)\, dt$ ! – Mercy King Dec 13 '12 at 18:43
@Mercy: but this is the same. – Mhenni Benghorbal Dec 13 '12 at 19:14
It is not because the integration is on a segment, not on the whole $\mathbb{R}$. – Mercy King Dec 13 '12 at 19:47

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