# Solve Rudin Theorem 11.35: Norm Inequality

• ehrenfest
In summary, Rudin uses the inequality 0 \leq \int_X (|f| + \lambda |g|)^2 d\mu = ||f||^2 + 2 \lambda \int_X |fg| d\mu + \lambda^2 ||g||^2 to show that \int_X |fg|d\mu \leq ||f|| ||g||. This can be achieved by setting lambda = -||f||/||g||, assuming that ||f|| and ||g|| are not zero.
ehrenfest
[SOLVED] Rudin Theorem 11.35

## Homework Statement

Rudin wants to show that

$$\int_X |fg|d\mu \leq ||f|| ||g||$$

He says that it follows from the inequality

$$0 \leq \int_X (|f| + \lambda |g|)^2 d\mu = ||f||^2 + 2 \lambda \int_X |fg| d\mu + \lambda^2 ||g||^2$$

which holds for any real lambda. Is there a special value of lambda that you are supposed to plug in here or is there a combination of values of lambda that I should use or should I take the limit as lambda goes to 0 or what...? I tried lambda = -1/2 but that didn't help at all. I don't see how you can get the norm of f and the norm of g multiplied instead of added

## The Attempt at a Solution

Have you tried to set lambda = something involving ||f|| and ||g||?

morphism said:
Have you tried to set lambda = something involving ||f|| and ||g||?

we can assume that ||f|| and ||g|| are not zero since then f and g are zero except on a set of measure zero, so the integral must be zero, so we set lambda equal to -||f||/||g|| and the theorem follows, thanks.

1.

## What is Rudin Theorem 11.35 and why is it important in mathematics?

Rudin Theorem 11.35, also known as the Norm Inequality, is a fundamental result in functional analysis that relates the norm of a linear operator to the norm of its adjoint. It is important in mathematics because it has many applications in various fields, including analysis, differential equations, and mathematical physics.

2.

## What is the statement of Rudin Theorem 11.35?

The statement of Rudin Theorem 11.35 is as follows: Let X and Y be normed spaces and T be a bounded linear operator from X to Y. Then, the norm of the adjoint of T is less than or equal to the norm of T, i.e. ||T*|| ≤ ||T||.

3.

## How is Rudin Theorem 11.35 proven?

Rudin Theorem 11.35 can be proven using the properties of the norm and the definition of the adjoint operator. One approach is to show that for any x in X, ||T*(x)|| ≤ ||T||*||x||, and then use the definition of the norm of an operator to conclude that ||T*|| ≤ ||T||.

4.

## What are some examples of applications of Rudin Theorem 11.35?

Rudin Theorem 11.35 is used in many areas of mathematics, including functional analysis, differential equations, and mathematical physics. For example, it is used in the study of differential equations to prove existence and uniqueness of solutions, and in mathematical physics to analyze the behavior of physical systems.

5.

## Are there any generalizations or extensions of Rudin Theorem 11.35?

Yes, there are several generalizations and extensions of Rudin Theorem 11.35. Some of these include the norm inequality for bounded operators between Banach spaces, the norm inequality for Hilbert spaces, and the norm inequality for unbounded operators. These generalizations and extensions have important applications in various fields of mathematics and physics.

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