Question

Problem. In Example 1 and Example 4 of the slide, we introduced two norms in C[-1, 1) and explained the completeness of C[-1,1] with respect to each norms. In this problem, we set Illu||| := |u(x)|dr. ||$|| := sup|u(x)]. RE(-1,1) Then, two norms, ||| . ||| and || . ||, are equivalent or not? (Equivalence of norms were introduced in the second lecture of this course.) If that assertion is true, then please write “True.” and prove it. If that assertion is not true, then please write "False." and explain by giving an example. At least, please write "True." or "False." before the deadline.

Answer 1

False :

Let us define $u(x)=\begin{cases} 1+x&-\frac{1}{n}\leq x\leq 0\\1-x&0\leq x\leq \frac{1}{n}\\0&\text{otherwise}\end{cases}$ .

we can see that $\|u\|_1=\sup_{x\in [-1,1]}|u(x)|=1$ because the function have the peak at $x=0$ .

Since this function forms a triangle with the base length $b=\frac{2}{n}$ and height $h=1$ , its area area is $\|u\|_2=\int_{-1}^1|u(x)|\;dx=\frac{1}{2}bh=\frac{1}{n}$ .

Since there is no for all n such that $1\leq C\frac{1}{n}$ , i.e., there is C for all n such that $\|u\|_1\leq C\|u\|_2$ . Hence these two norms are not equivalent.