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Problem. Let $1\le p<\infty$ and $n \ge 1$ and let $W^n_p[0,1]=$ the functions $f:[0,1]\to \Bbb{C}$ such that $f$ has $n-1$ continuous derivatives, $f^{(n-1)}$ is absolutely continuous, and $f^{(n)} \in L^p[0,1]$. For $f$ in $W^n_p[0,1]$, define $$||f||:=\sum_{k=0}^n\left[\int_0^1|f^{(k)}(x)|^p ~\mathrm{d}x\right]^\frac{1}{p}$$ Then $W^n_p[0,1]$ is a Banach space.

If we take a Cauchy sequence $(f_j)_j$ in $W^n_p[0,1]$, then it follows that $(f_j^{(k)})_j$ is Cauchy in $L^p[0,1]$ for all $k = 0, 1, \dots, n$. Hence there are $g_{k} \in L^p[0,1]$ such that $(f_j^{(k)})\to g_k$ for $k = 0, 1, \dots, n$. From this how to prove that $g_k=g_0^{(k)}$ for all $k = 1, \dots, n$? Thanks.

EDIT I tried to proceed as follows: $f_j \to g_0$ in $L^p$ $\implies \text{for some subsequence }f_{j_l}\to g_0$ a.e. on $[0,1]$. Then we have $$g_0(x)= \lim f_{j_l}(x)= \lim \int_0^x f_{j_l}^{'}=\int_0^x\lim f_{j_l}^{'}=\int_0^x g_1$$ for a.e $x$ in $[0, 1]$. This implies $g_0'(x) = g_1(x)$ a.e. $x$ . Now similarly we can prove others. Then since $f_j^{(n-1)}$ was absolutely continuous functions with $L^p$ derivatives we can use this to prove that $g_{n-1}(=g_0^{(n-1)})$ is also absolutely continuous with $L^p$ derivative. And this shows that $g_0 \in W^n_p[0,1]$.

Is there any mistake?

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    $\begingroup$ Does this answer your question? Space Sobolev $W^{m,p}$ complete $\endgroup$
    – lc2r43
    Commented May 13, 2021 at 6:04
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    $\begingroup$ @lc2r43 No. Actually I don't know "weak derivatives". $\endgroup$
    – sigma
    Commented May 13, 2021 at 6:07
  • $\begingroup$ How do you define the derivative $f^{(n)}$ if not as weak derivative? $\endgroup$
    – daw
    Commented May 14, 2021 at 11:33
  • $\begingroup$ @daw I would define it as : $f^{(n)}(x):=\lim_{h \to 0}\frac{f^{(n-1)}(x+h)-f^{(n-1)}(x)}{h}$ for all $x$. Here $f^{(n-1)}$ is given to be differentiable with a $L^p$ derivative. Right?? $\endgroup$
    – sigma
    Commented May 15, 2021 at 3:20
  • $\begingroup$ No, that limit is not guaranteed to exist, and definitely not for all x, you should learn the definition of a weak derivative before coming back to this question $\endgroup$ Commented May 21, 2021 at 0:23

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