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Let $u_n ∈ W^{1,1} (I), I=(0,1)$ defined by:

$u'_n (x) = n$ if $x < 1/n$

$u'_n (x) = 0$ if $x > 1/n$

$u_n (0) = 0$.

Find $||u_n − 1||_∞$.

My attempt:

We have $u_n(x)=u_n(0)+\int_0^{1/n}ndt=1$ a.e. hence $||u_n − 1||_∞=0$ but there is a case I can't really grasp is when $n=\infty$ therefore $||u_n − 1||_∞=1$ so what is the correct answer?

Edit

Correcting my dumb mistake thanks to comments.

if $x<1/n$ , we have $u_n(x)=u_n(0)+\int_0^{x}n\ dt$ hence $||u_n − 1||_∞=1$ as $x\to 0$ when $n\to \infty$.

if $x>1/n$ we have $u_n(x)=u_n(0)+\int_0^{1/n}n\ dt=1$ hence $||u_n − 1||_∞=||1-1||_∞=0$.

We deduce finally that $||u_n − 1||_∞=1$

is That correct?

Thank you!

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    $\begingroup$ Your expression for $u_n(x)$ doesn't have $x$ in it. Try to split into cases carefully $\endgroup$ – Calvin Khor May 5 at 16:34
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    $\begingroup$ $u_n(x) = u_n(0) + \int_0^x u_n'(t) dt$. This is different from what you wrote if $x < 1/n$. Note for example that $\|u_n - 1\|_\infty \geq |u_n(0) - 1| = 1$. $\endgroup$ – Rhys Steele May 5 at 16:34
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    $\begingroup$ You are almost correct but you do not understand the notation correctly. Note carefully $$ \|u_n - 1\|_\infty = \sup_{x\in(0,1)} |u_n(x) - 1|.$$ Here $x$ is a bound variable, you can't send $x\to 0$. $\endgroup$ – Calvin Khor May 5 at 18:38

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