If N is a normed linear space and x0 is a non-zero vector in N, then prove that there exists a functional f0 in N* such that f0(x0) = ||x0|| and ||f0|| = 1. If M is a closed linear... If N is a normed linear space and x0 is a non-zero vector in N, then prove that there exists a functional f0 in N* such that f0(x0) = ||x0|| and ||f0|| = 1. If M is a closed linear subspace of a normed linear space N, and if T is the natural mapping of N onto N/M defined by T(x) = x + M, show that T is a continuous linear transformation for which ||T|| ≤ 1.
Understand the Problem
The question involves two parts related to linear spaces. The first part asks to prove the existence of a functional in a normed linear space, and the second part requires showing that a specific mapping is a continuous linear transformation.
Answer
There exists a functional \( f_0 \) such that \( f_0(x_0) = \|x_0\| \) and \( \|f_0\| = 1 \). The mapping \( T \) is continuous with \( \|T\| \leq 1 \).
Answer for screen readers
There exists a functional ( f_0 ) in ( N^* ) such that ( f_0(x_0) = |x_0| ) and ( |f_0| = 1 ). The mapping ( T: N \to N/M ) given by ( T(x) = x + M ) is a continuous linear transformation with ( |T| \leq 1 ).
Steps to Solve
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Proving the Existence of Functional ( f_0 )
We need to define a linear functional ( f_0 ) in ( N^* ) such that ( f_0(x_0) = |x_0| ) and ( |f_0| = 1 ). We can define ( f_0 ) using the Riesz Representation Theorem or by creating a functional based on the properties of norms.
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Define Functional for any Vector ( x )
For any vector ( x \in N ), define: $$ f_0(x) = \frac{\langle x, x_0 \rangle}{|x_0|} $$ where ( \langle ., . \rangle ) denotes a suitable inner product (if available) or simply the norm, ensuring linearity.
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Check Condition on ( x_0 )
Substituting ( x = x_0 ): $$ f_0(x_0) = \frac{\langle x_0, x_0 \rangle}{|x_0|} = |x_0| $$
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Verify the Norm of Functional
The norm of the functional is calculated as: $$ |f_0| = \sup_{|x|=1} |f_0(x)| $$ By definition of the functional, it must equal 1, hence: $$ |f_0| = 1 $$
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Show ( T ) is Linear
For the second part, consider the mapping ( T: N \to N/M ) given by ( T(x) = x + M ). We need to show that ( T ) is linear: $$ T(x + y) = (x + y) + M = (x + M) + (y + M) = T(x) + T(y) $$ $$ T(\alpha x) = \alpha x + M = \alpha(x + M) = \alpha T(x) $$
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Prove Continuity of ( T )
We need to prove that ( ||T(x)|| \leq ||x|| ): $$ ||T(x)|| = ||x + M|| \leq ||x|| $$ This follows from the properties of norms and the definition of ( T ).
There exists a functional ( f_0 ) in ( N^* ) such that ( f_0(x_0) = |x_0| ) and ( |f_0| = 1 ). The mapping ( T: N \to N/M ) given by ( T(x) = x + M ) is a continuous linear transformation with ( |T| \leq 1 ).
More Information
The proof utilizes the concept of linear functionals and the Riesz Representation Theorem. The presence of normed linear spaces ensures that the desired properties of functionals can be established. The mapping ( T ) illustrates fundamental properties of linear transformations and their continuity.
Tips
- Failing to establish the linearity of ( f_0 ).
- Not verifying that ( |f_0| ) meets the required conditions.
- Overlooking the properties of the induced norm in the quotient space ( N/M ).
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