Equivalent Norms: Piecewise Continuous Linear Function [0,1]

[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?

 

[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

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

 

[antiemptyv]

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

 

[morphism]

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

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

 

[Pere Callahan]

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

That function is certainly piecewise continuous 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.