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

**oztuserd** · March 3rd 2009, 01:23 PM

I could not prove this.

**Opalg** · March 4th 2009, 01:17 PM

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

**oztuserd** · March 4th 2009, 02:04 PM

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

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

**Opalg** · March 5th 2009, 12:43 AM

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

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

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

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

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

need to show violation of sclar multiplication

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