Innner product for which derivative operator is bounded

In summary, the conversation discusses the boundedness of the derivative operator d/dx with respect to a specific norm given by an inner product. The inner product is defined for functions in the space C^1([0,1]) and the corresponding norms are given by the integral of the function and its derivative. The conversation concludes that the derivative operator is indeed bounded with respect to these norms.
  • #1
esisk
44
0
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)
 
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  • #2
Note that [tex]\frac{d}{dx} : C^1([0,1]) \rightarrow C^0([0,1])[/tex] you defined the inner product (not the norm) on [tex]C^1[/tex], but what is that of [tex]C^0[/tex]? I'm going to assume you give it the inner product: [tex]<f,g> = \int_{0} ^{1} f\overline{g}[/tex]. Then the norms induced by these are [tex]\Vert f \Vert _{C^1} = \left( \int_{0}^{1} \vert f \vert ^2 + \int_{0}^{1} \vert f' \vert ^2 \right) ^{1/2}[/tex] and [tex]\Vert g \Vert _{C^0} = \left( \int_{0}^{1} \vert g \vert ^2 \right) ^{1/2}[/tex]

Then [tex]\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[/tex] so it's bounded with respect to those norms.
 
  • #3
Jose,
Thank you very much.
Regards
 

1. What is an inner product?

An inner product is a mathematical operation that takes in two vectors and produces a scalar value. It is similar to the dot product, but it allows for more general vector spaces and has additional properties such as symmetry and linearity.

2. How is the derivative operator related to inner product?

The derivative operator is related to inner product through the concept of a bounded operator. A bounded operator is one that does not grow too quickly, and the derivative operator is an example of this. In particular, the derivative operator is bounded on a space of differentiable functions when using the inner product defined by integrating the product of two functions.

3. What does it mean for a derivative operator to be bounded?

When a derivative operator is bounded, it means that the operator does not cause the functions it operates on to grow too quickly. This is important because it allows for certain properties and theorems to hold, and it allows for the use of techniques such as the Banach-Steinhaus theorem.

4. What are some applications of inner product for which the derivative operator is bounded?

One application of this concept is in the study of differential equations. By using the inner product and the boundedness of the derivative operator, one can prove existence and uniqueness theorems for solutions to certain types of differential equations. This is also useful in optimization problems and in functional analysis.

5. How is the boundedness of the derivative operator proven?

The boundedness of the derivative operator is typically proven using techniques from functional analysis, such as the Banach-Steinhaus theorem or the Hahn-Banach theorem. These theorems provide conditions under which a linear operator, such as the derivative operator, is bounded. Additionally, specific properties of the inner product and the space in which it is defined can also be used to prove boundedness.

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