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Let $(f_n)_{n\ge 1}$ be a sequence of function that is differentiable. such that $f_n\xrightarrow{L^p}f$ converges. I was wondering if $f_n'\xrightarrow{L^p}f'$. So far I could only find this result but it doesn't really apply here I think.

edamondo
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    Of course not. If $f$ is measurable and non continuous, there is a sequence of $\mathcal C^\infty $ function $(f_n)$ s.t. $f_n\to f$ in $L^p$. So, if $f_n\to f$ in $L^p$ and $(f_n)$ differentiable, there is no reason for $f$ to be differntiable. – Surb Jan 14 '21 at 13:37

1 Answers1

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No.

Example: $p=1$, $f_n$ and $f$ in $L^1[0,1]$ let be given by

$$ f_n(x) =x^n , \quad f(x)=0.$$

Then

$$||f_n-f||_1 = \frac{1}{n+1} \to 0,$$

as $ n \to \infty,$ but

$$||f_n'-f'||_1 = 1$$

for all $n$.

Fred
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