not Show that ([0,1]_is in the l' metric d (fig)= (51f6)-96) ³d x) ² fumpluts
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Topic: Completeness in Metric Spaces**
**Transcription of Handwritten Notes:**
"Show that \( C[0,1] \) is not complete in the \( L^1 \) metric.
\[ d(f,g) = \left( \int_{0}^{1} |f(x) - g(x)|^2 \, dx \right)^{\frac{1}{2}} \]"
**Explanation:**
This note challenges the reader to demonstrate that the space of continuous functions on the interval \([0,1]\), denoted as \( C[0,1] \), lacks completeness when equipped with the \( L^1 \) metric. However, the given formula represents the \( L^2 \) metric (Euclidean norm for functions), suggesting a potential mix-up or further exploration in metric types.
**Mathematical Context:**
1. **Function Space \( C[0,1] \):** This space consists of all continuous functions defined on the interval \([0,1]\).
2. **Metric \( d(f,g) \):** The given metric is the \( L^2 \) metric derived from the integral of the squared differences between functions \( f \) and \( g \). It calculates the "distance" between two functions as the square root of the integral of the square of their difference.
3. **Completeness:** A metric space is complete if every Cauchy sequence of functions in the space converges to a limit that is also within the space.
Readers are invited to further explore the distinctions between \( L^1 \) and \( L^2 \) metrics and delve into why completeness may or may not hold in these contexts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F8ab7dfc7-bf60-48ca-8664-600e7b4adbe2%2Fnbx6um7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Completeness in Metric Spaces**
**Transcription of Handwritten Notes:**
"Show that \( C[0,1] \) is not complete in the \( L^1 \) metric.
\[ d(f,g) = \left( \int_{0}^{1} |f(x) - g(x)|^2 \, dx \right)^{\frac{1}{2}} \]"
**Explanation:**
This note challenges the reader to demonstrate that the space of continuous functions on the interval \([0,1]\), denoted as \( C[0,1] \), lacks completeness when equipped with the \( L^1 \) metric. However, the given formula represents the \( L^2 \) metric (Euclidean norm for functions), suggesting a potential mix-up or further exploration in metric types.
**Mathematical Context:**
1. **Function Space \( C[0,1] \):** This space consists of all continuous functions defined on the interval \([0,1]\).
2. **Metric \( d(f,g) \):** The given metric is the \( L^2 \) metric derived from the integral of the squared differences between functions \( f \) and \( g \). It calculates the "distance" between two functions as the square root of the integral of the square of their difference.
3. **Completeness:** A metric space is complete if every Cauchy sequence of functions in the space converges to a limit that is also within the space.
Readers are invited to further explore the distinctions between \( L^1 \) and \( L^2 \) metrics and delve into why completeness may or may not hold in these contexts.
Expert Solution
Step 1
The given set is X = C[0,1] and the metric defined on X is which is defined as
We have to show that C[0,1] is not complete with this
Step by step
Solved in 2 steps with 1 images
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