Innner product for which derivative operator is bounded

[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)

 

[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: &lt;f,g&gt; = \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&#039; \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&#039; \Vert _{C^0} ^2 = \int_{0}^{1} \vert f&#039; \vert ^2 \leq \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f&#039; \vert ^2 = \Vert f \Vert _{C^1} ^2 so it's bounded with respect to those norms.

 

[esisk]

Jose,
Thank you very much.
Regards