I have $ f\in L^1(0,1)$ such that $ xf$ is absolutely continuous on $ [0,1]$ . The latter is equivalent to $ f’\in L^1_{\rm loc}(0,1)$ and $ xf’\in L^1(0,1)$ .

Does it follow from these assumptions that $ (xf)(0) = 0$ ?

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# Tag: $L^101$

## $f\in L^1(0,1)$, $f’\in L^1_{\rm loc}(0,1)$ exists, $xf’\in L^1(0,1)$. Does then $(xf)(0+) = 0$?

## Convergence of a recursion in $L^1(0,1)$

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I have $ f\in L^1(0,1)$ such that $ xf$ is absolutely continuous on $ [0,1]$ . The latter is equivalent to $ f’\in L^1_{\rm loc}(0,1)$ and $ xf’\in L^1(0,1)$ .

Does it follow from these assumptions that $ (xf)(0) = 0$ ?

Let $ \mu_0>0$ , $ a>0$ , $ b>0$ , and $ f(t)$ , $ g(t)>0$ , $ p(t)$ be some continuously differentiable functions over $ \mathbb{R}$ .

I am looking for various tools to study the stability of the following recursive formula $ $ h_n(t)=\frac{-\mu_{n-1} f'(t)}{g(\mu_{n-1} f(t))}, \; \forall t\in [0,1],$ $

$ $ \mu_n=\frac{1}{a}\int_0^1h_n(t)\int_{t}^1p\left(b – \int_{0}^{s}h_n(r)dr\right)dsdt.$ $

I specifically want to know if $ h_n(t)$ is convergent in $ L^1(0,1)$ .

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