(#12) Show 0 € int(S) implies Sº is bounded.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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This is from an optimization and convex analysis class

**Problem #12: Boundedness of the Polar Set**

Show that if \( 0 \in \text{int}(S) \) implies that \( S^\circ \) is bounded.

In this problem, we are asked to demonstrate that if the origin \( 0 \) is an interior point of the set \( S \), then the polar set \( S^\circ \) is bounded. This involves concepts from convex analysis and functional analysis, particularly concerning interior points and polar sets. Here’s an outline of the principles involved:

1. **Interior Points**: The notation \( 0 \in \text{int}(S) \) signifies that there exists some neighborhood around the origin that is completely contained within the set \( S \). This property is essential for showing the boundedness of the polar set.

2. **Polar Set**: The polar of a set \( S \), denoted \( S^\circ \), is defined as the collection of all vectors \( y \) such that the inner product \( \langle x, y \rangle \leq 1 \) for all \( x \) in \( S \). The concept of a polar set is important in duality theories and optimization.

3. **Boundedness**: A set is bounded if there exists some positive number \( M \) such that the norm of every element in the set is less than or equal to \( M \). Here, we show that \( S^\circ \) is bounded given the interior point condition.

To prove this, you would typically leverage the separation theorems and the properties of convex sets to demonstrate that if the origin is an interior point of \( S \), then there is a radius within which the properties of \( S \) can be exploited to show the boundedness of \( S^\circ \). A detailed step-by-step proof would be covered in a lecture or textbook on advanced convex analysis.
Transcribed Image Text:**Problem #12: Boundedness of the Polar Set** Show that if \( 0 \in \text{int}(S) \) implies that \( S^\circ \) is bounded. In this problem, we are asked to demonstrate that if the origin \( 0 \) is an interior point of the set \( S \), then the polar set \( S^\circ \) is bounded. This involves concepts from convex analysis and functional analysis, particularly concerning interior points and polar sets. Here’s an outline of the principles involved: 1. **Interior Points**: The notation \( 0 \in \text{int}(S) \) signifies that there exists some neighborhood around the origin that is completely contained within the set \( S \). This property is essential for showing the boundedness of the polar set. 2. **Polar Set**: The polar of a set \( S \), denoted \( S^\circ \), is defined as the collection of all vectors \( y \) such that the inner product \( \langle x, y \rangle \leq 1 \) for all \( x \) in \( S \). The concept of a polar set is important in duality theories and optimization. 3. **Boundedness**: A set is bounded if there exists some positive number \( M \) such that the norm of every element in the set is less than or equal to \( M \). Here, we show that \( S^\circ \) is bounded given the interior point condition. To prove this, you would typically leverage the separation theorems and the properties of convex sets to demonstrate that if the origin is an interior point of \( S \), then there is a radius within which the properties of \( S \) can be exploited to show the boundedness of \( S^\circ \). A detailed step-by-step proof would be covered in a lecture or textbook on advanced convex analysis.
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