Innner product for which derivative operator is bounded

1. Nov 13, 2009

esisk

Is the derivative operator d/dx bounded with respect to the norm <f,g> defined by

integral from 0 to 1 of f g* +f'g*'

where * denotes conjugation. Thank you. (Not homework)

2. Nov 13, 2009

Jose27

Note that $$\frac{d}{dx} : C^1([0,1]) \rightarrow C^0([0,1])$$ you defined the inner product (not the norm) on $$C^1$$, but what is that of $$C^0$$? I'm going to assume you give it the inner product: $$<f,g> = \int_{0} ^{1} f\overline{g}$$. Then the norms induced by these are $$\Vert f \Vert _{C^1} = \left( \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f' \vert ^2 \right) ^{1/2}$$ and $$\Vert g \Vert _{C^0} = \left( \int_{0}^{1} \vert g \vert ^2 \right) ^{1/2}$$

Then $$\Vert f' \Vert _{C^0} ^2 = \int_{0}^{1} \vert f' \vert ^2 \leq \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f' \vert ^2 = \Vert f \Vert _{C^1} ^2$$ so it's bounded with respect to those norms.

3. Nov 14, 2009

esisk

Jose,
Thank you very much.
Regards