Chapter 5: Problem 13
A function \(f\) is said to be symmetrically continuous at a point \(x\) if $$ \lim _{h \rightarrow 0}[f(x+h)-f(x-h)]=0 $$ Show that if \(f\) is continuous at a point, then it must be symmetrically continuous there and that the converse does not hold.
Short Answer
Expert verified
If a function is continuous, it is symmetrically continuous; however, symmetric continuity doesn't imply continuity, as shown by counterexamples.
Step by step solution
01
Identifying Given Conditions
We are given that the function \(f\) is continuous at a point \(x\). By the definition of continuity, for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(|h| < \delta\), then \(|f(x+h) - f(x)| < \varepsilon\). We need to prove that this implies symmetric continuity at \(x\) as well.
02
Applying Symmetric Continuity Definition
We want to prove that \(\lim _{h \rightarrow 0}[f(x+h)-f(x-h)]=0\). According to this expression, for symmetric continuity, both \(|f(x+h) - f(x)|\) and \(|f(x-h) - f(x)|\) are less than \(\varepsilon\) when \(h\) is sufficiently small. Thus \(|f(x+h) - f(x-h)| < 2\varepsilon\).
03
Deriving Symmetric Continuity from Continuity
Given \(|f(x+h) - f(x)| < \varepsilon\) and \(|f(x-h) - f(x)| < \varepsilon\), we apply the triangle inequality to get: \[|f(x+h) - f(x-h)| \leq |f(x+h) - f(x)| + |f(x) - f(x-h)| < \varepsilon + \varepsilon = 2\varepsilon.\] This implies that \(\lim _{h \rightarrow 0} [f(x+h) - f(x-h)] = 0\), proving symmetric continuity.
04
Constructing a Counterexample for the Converse
To show that the converse does not hold, consider the function \(f(x)\) defined by \(f(x) = x\) for \(x eq 0\) and \(f(0) = 1\). At \(x = 0\), for any \(h eq 0\), \(f(h) - f(-h) = h - (-h) = 2h\), and thus \(\lim _{h \rightarrow 0}[f(h) - f(-h)] = \lim _{h \rightarrow 0} 2h = 0\) showing symmetry continuity. However, \(f\) is not continuous at \(0\) as \(\lim_{x \to 0} f(x) = 0\) which is not equal to \(f(0) = 1\).
05
Conclusion of Analysis
We demonstrated that a continuous function at a point is also symmetrically continuous at that point by using the definition of continuity and applying the triangle inequality. Subsequently, we constructed a specific counterexample, proving that a function can be symmetrically continuous at a point without being continuous there.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
Continuity is a central idea in calculus and real analysis, expressing the intuitive notion that small changes in the input of a function result in small changes in the output. In mathematical terms, we define continuity at a point as follows. A function \( f \) is continuous at a point \( x \) if, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |h| < \delta \), it follows that \( |f(x+h) - f(x)| < \varepsilon \).
Here’s what this means in simpler words:
Here’s what this means in simpler words:
- \( \varepsilon \) represents any small positive number that we choose; it's like a target accuracy for how close the function's output should stay close to \( f(x) \).
- \( \delta \) corresponds to a small range around \( x \) such that whenever we choose an input \( x+h \) or \( x-h \) within this range, the resulting output \( f(x+h) \) is within \( \varepsilon \) of \( f(x) \).
Limit
A limit describes the behavior of a function as the input approaches a specific value, often giving insight into the function's continuity. For symmetric continuity, the limit definition is pivotal, defined as follows: the symmetric limit at a point \( x \) is described by \( \lim_{h \rightarrow 0} [f(x+h) - f(x-h)] = 0 \).
Understanding limits involves:
Understanding limits involves:
- Approaching a point: Limits essentially ask what happens as we get closer and closer to a certain \( x \.\) They don't depend on actually reaching \( x \), just getting infinitely close.
- Behavior over exactness: Limits focus on what values a function approaches rather than the exact spot it lands.
Triangle Inequality
The triangle inequality is a fundamental mathematical principle often employed to derive symmetric continuity from standard continuity. It states that for real numbers (or vectors) \( a \), \( b \), the relationship \( |a + b| \leq |a| + |b| \) always holds.
This concept assists in:
This concept assists in:
- Bounding expressions: By splitting the left and right deviations \(|f(x+h) - f(x)|\) and \(|f(x-h) - f(x)|\), and adding them to form \(|f(x+h) - f(x-h)|\), we can establish a combined bound.
- Ensuring Sum Control: By using \(|f(x+h) - f(x-h)| < 2\varepsilon\), the total change between '\( x+h \)' and '\( x-h \)' can be made arbitrarily small.
Counterexample
A counterexample is a powerful tool to demonstrate the limitations or boundaries of a mathematical proposition. Despite continuous functions at a point ensuring symmetric continuity, the reverse is not always true, which can be shown using a counterexample.
Consider a function \( f \) defined such that \( f(x) = x \) for all \( x eq 0 \) and \( f(0) = 1 \). To understand why this is significant:
Consider a function \( f \) defined such that \( f(x) = x \) for all \( x eq 0 \) and \( f(0) = 1 \). To understand why this is significant:
- Symmetric Continuity Achieved: At \( x=0 \), for small \( h eq 0 \), we have \( f(h) - f(-h) = h - (-h) = 2h \) and as \( h \rightarrow 0 \), this approaches zero. Hence, it is symmetrically continuous.
- Lack of Regular Continuity: The limit of \( f(x) \) as \( x \rightarrow 0 \) is 0 (since \( f \) approaches the identity function), distinct from \( f(0) = 1 \); thus, it is not continuous at \( x=0 \).