Given $u\in W^{1,p}([0,L])$, for each $n \in \Bbb N$ , split the interval $[0,L]$ into $n+1$ sub intervals of the form $[x_i,x_{i+1}]$ where $x_i = \dfrac{iL}{n}$
and define for each $n $
$u_n:[x_i,x_{i+1}] \to \Bbb R$ by $u_n(x) = u(x_i)+\dfrac{n(u(x_{i+1}-u(x_i))}{L}(x-x_i)$
That is, $u_n$ is the "lines" connecting $u(x_i)$ to $u(x_{i+1})$.
I'm trying to show that $u_n\to u$ in $L^p(0,L)$ but with no success , I appreciate any help.
Thanks!
Here is an elementary proof for $p\in (1,\infty)$. More advanced arguments can be found in textbooks on finite element methods.
Let $e:= u_n-u$. Then $e(x_i)=0$ for all $i$. By the fundamental theorem, we get (with $p>1$) $$ \int_{x_i}^{x_{i+1}} |e(x)|^p dx = \int_{x_i}^{x_{i+1}} \int_{x_i}^x p e'(s)sign(e(s))|e(s)|^{p-1} ds \ dx. $$ The inner integral of right-hand side can be estimated by $$ \int_{x_i}^x e'(s)sign(e(s))e(s)^{p-1} ds \le \int_{x_i}^{x_{i+1}} |e'(s)|\cdot |e(s)|^{p-1}ds $$ Summing these integrals gives with $h:=L/n=x_{i+1}-x_i$ $$ \|e\|_p^p \le h\int_0^L |e'(s)|\cdot |e(s)|^{p-1}ds \le\left( \int_0^L |e'(s)|^pds \right)^{1/p} \left( \int_0^L |e(s)|^pds \right)^{(p-1)/p} = h \|e'\|_{p} \|e\|_p^{p-1}, $$ which proves $\|e\|_p\le h \|e'\|_{p}$.
It remains to show that $\|e'\|_{p}$ is bounded wrt $n$. We have $e'(x) = h^{-1} ( u(x_{i+1}-u(x_i)) - u'(x)$ for $x\in (x_i,x_{i+1})$. We know $u'\in L^p$, so it remains to investigate the difference quotient contribution: $$ \int_{x_i}^{x_{i+1}} |h^{-1} ( u(x_{i+1})-u(x_i))|^p = h^{1-p} |u(x_{i+1})-u(x_i)|^p $$ and $$ |u(x_{i+1})-u(x_i)| \le \int_{x_i}^{x_{i+1}} |u'(s)|ds \le \left(\int_{x_i}^{x_{i+1}} |u'(s)|^pds\right)^{1/p}\cdot h^{(p-1)/p}, $$ so $$ \int_{x_i}^{x_{i+1}} |h^{-1} ( u(x_{i+1})-u(x_i))|^p \le h^{1-p} \int_{x_i}^{x_{i+1}} |u'(s)|^pds\cdot h^{(p-1)} $$ and $\|e'\|_p \le 2 \|u'\|_p$.