# Thread: Show that ||x|| is not a norm on Lp[0,1]

1. ## Show that ||x|| is not a norm on Lp[0,1]

I could not prove this.

2. Let f be the function that is equal to 1 on the interval [0,1/2] and 0 on the interval [1/2,1], and let g=1-f.

Then $\displaystyle \|f\|_p = \|g\|_p = (1/2)^{1/p}<1/2$, and $\displaystyle \|f+g\|_p = 1$. So $\displaystyle \|f+g\|_p>\|f\|_p+\|g\|_p$.

3. ## need to show violation of sclar multiplication

Opalq, thank you for your response. That is a very good proof. But I need to use the hint given in the question. That is I need to show the violation of

$\displaystyle \|cx\|_p=|c| \|x\|_p$ where c is a scalar, 0 < c < 1

4. Originally Posted by oztuserd
Opalq, thank you for your response. That is a very good proof. But I need to use the hint given in the question. That is I need to show the violation of

$\displaystyle \|cx\|_p=|c| \|x\|_p$ where c is a scalar, 0 < c < 1
I think you are misunderstanding the hint. You will never find a function x such that $\displaystyle \|cx\|_p \ne |c| \|x\|_p$, because $\displaystyle \|cx\|_p = \left[\int_0^1|cx(t)|^pdt\right]^{1/p} = \left[|c|^p\int_0^1|x(t)|^pdt\right]^{1/p} = |c|\|x\|_p$.

In fact, I used the hint in my example, in the inequality $\displaystyle (1/2)^{1/p}<1/2$, which is equivalent to $\displaystyle 1/2<(1/2)^p$.