$||.||_{\infty}$ in sobolev space

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!

• Your expression for $u_n(x)$ doesn't have $x$ in it. Try to split into cases carefully – Calvin Khor May 5 at 16:34
• $u_n(x) = u_n(0) + \int_0^x u_n'(t) dt$. This is different from what you wrote if $x < 1/n$. Note for example that $\|u_n - 1\|_\infty \geq |u_n(0) - 1| = 1$. – Rhys Steele May 5 at 16:34
• 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 $x\to 0$. – Calvin Khor May 5 at 18:38