Problem 3
Question
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a differentiable function such that \(f^{\prime}\) is a bounded function. Prove \(f\) is a Lipschitz continuous function.
Step-by-Step Solution
Verified Answer
Since \(|f'(x)| \leq M\), \(f\) is Lipschitz continuous with constant \(M\).
1Step 1: Understand the Definition of Bounded Derivative
A function is said to have a bounded derivative if there exists some constant \(M > 0\) such that \(|f'(x)| \leq M\) for all \(x \in \mathbb{R}\). This means the rate of change of \(f(x)\) is limited by \(M\) at any point along the real line.
2Step 2: Recall the Meaning of Lipschitz Continuity
A function \(f\) is Lipschitz continuous if there exists a constant \(L\geq 0\) such that for all \(x, y \in \mathbb{R}\), \(|f(x) - f(y)| \leq L|x-y|\). The smallest such \(L\) is called the Lipschitz constant.
3Step 3: Apply the Mean Value Theorem
The Mean Value Theorem (MVT) states that if \(f\) is differentiable on \((a, b)\), then there exists some \(c \in (a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). This implies that \(|f(b) - f(a)| = |f'(c)||b-a|\).
4Step 4: Use Bounded Derivative to Prove Lipschitz Condition
Since \(|f'(x)| \leq M\) for all \(x\), from MVT, we have \(|f(b) - f(a)| = |f'(c)||b-a| \leq M|b-a|\). This shows that \(f\) is Lipschitz continuous with \(L = M\).
5Step 5: Conclude the Result
We have shown that the bounded derivative condition implies the existence of a constant \(M\) such that for any two real numbers \(a\) and \(b\), \(|f(a) - f(b)| \leq M|a-b|\). Thus, \(f\) is a Lipschitz continuous function.
Key Concepts
Bounded DerivativeMean Value TheoremDifferentiable Function
Bounded Derivative
A bounded derivative means that the derivative, or the instantaneous rate of change, of a function does not exceed a certain constant. This is crucial because it controls how sharply or steeply a function can ascend or descend. To say a function has a bounded derivative is to say there exists a value \( M > 0 \) such that for every point \( x \) in the function's domain, the absolute value of the derivative, \( |f'(x)| \), is less than or equal to \( M \). This implies:
- The function is not changing too rapidly at any point.
- There are no extreme spikes or valleys without limit.
- Its graph is smooth and does not have breaks or sharp turns.
Mean Value Theorem
The Mean Value Theorem (MVT) is a powerful tool in calculus that connects the derivative of a function with its overall change over an interval. It states that for a function \( f \) which is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists some point \( c \) in \( (a, b) \) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]This theorem essentially says that somewhere between any two points \( a \) and \( b \), the function must have a tangent that is parallel to the secant joining them. Key insights include:
- MVT assures us that the average rate of change is realized at some point within the interval.
- This equivalently means that \(|f(b) - f(a)|\) can be understood or estimated using \(|f'(c)|\times|b-a|\).
- It's a cornerstone in proving properties like Lipschitz continuity when combined with a bounded derivative.
Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. In essence, at each point \( x \), the function \( f \) can be locally approximated by a line (tangent), and this line changes smoothly without jumping or discontinuities. Differentiability entails:
- Continuity - the function has no gaps or jumps.
- Smoothness - the function does not have sharp corners (cusps).
- Predictability - knowing the function's behavior locally through its derivative.
Other exercises in this chapter
Problem 2
Suppose p is a polynomial of degree d. Given any \(x_{0} \in \mathbb{R}\), show that the \((d+1)\) th Taylor polynomial for \(p\) at \(x_{0}\) is equal to \(p\)
View solution Problem 2
Prove the quotient rule. Hint: You can do this directly, but it may be easier to find the derivative of \(1 / x\) and then use the chain rule and the product ru
View solution Problem 3
Let \(f(x):=|x|^{3}\). Compute \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) for all \(x\), but show that \(f^{(3)}(0)\) does not exist.
View solution Problem 3
For \(n \in \mathbb{Z}\), prove that \(x^{n}\) is differentiable and find the derivative, unless, of course, \(n
View solution