Equivalent norms?

1. Oct 17, 2008

antiemptyv

Suppose that ||f||= int 01| f(x) | dx and f is a piecewise continuous linear function on the interval [0,1]. If ||| f ||| = int 01 x | f(x) | dx, determine if the two norms are equivalent.

I know the first defines a norm, and the space is not complete. Can anyone offer any hints as to solving this problem?

2. Oct 17, 2008

Pere Callahan

Well, first of, when are two norms defined to be equivalent? If you know that, one part of the definition is very easily verified. For the other part consider the function

$$f_{\varepsilon}(x)=x\chi_{[0,\varepsilon]}(x)$$
What is then $\frac{||f_{\varepsilon}||}{|||f_{\varepsilon}|||}$? Conclusion?

Last edited: Oct 17, 2008
3. Oct 17, 2008

antiemptyv

i don't quite follow... is that function in the space?

4. Oct 17, 2008

morphism

$\chi_{[0,\varepsilon]}$ is the characteristic function of $[0,\varepsilon]$, i.e.

$$\chi_{[0,\varepsilon]}(x) = \begin{cases} 1 & \text{ if } x \in [0,\varepsilon] \\ 0 & \text{ if } x \in (\varepsilon, 1] \end{cases}.$$

5. Oct 17, 2008

Pere Callahan

That function is certainly piecewise continous and (piecewise) linear und thus in the space you mentioned.
Thanks for supplying this additional, crucial piece of information that by $\chi_A$ I meant the indicator function of A.