Question

• Prove that C.V100√n-1

Answer 1

We have to prove that $C^1[0,1]$ is not complete with the norm:

$||f||_{\infty}=sup_{x \in [0,1]}|f(x)|$

The right sequence for norm is $f_n=\sqrt{x+\dfrac{1}{x}}$

Notice that $n\in N:f_n\in C^1[0,1]$

let $f=\sqrt{x}$

We see that $f_n$ converges to f in sup norm in $C[0,1]$ , Thus it is cauchy.

$C^1[0,1]$ is a supspace of $C[0,1]$ and all terms of $(f_n)$ are in $C^1[0,1]$ ,So

As the $f_n$ converges to $C[0,1]$

$\implies (f_n) \text{ is Cauchy in } C^1[0,1].$

So Given space is not complete with the norm $||f(x)||_{\infty}=sup_{x\in[0,1]}|f(x)|$ .