Let f be a continuously differentiable function and let norm of Df be less than equal to some M. Prove that f is lipschitz continuous

To prove that a continuously differentiable function \( f: \mathbb{R}^n \rightarrow \mathbb{R}^m \) is Lipschitz continuous given that the norm of its derivative \( Df \) is bounded by some constant \( M \), we'll proceed with the following steps:

### 1. Statement of the Problem
Given:
- \( f \) is continuously differentiable.
- The norm of the derivative \( Df \) satisfies \( \|Df(x)\| \leq M \) for all \( x \in \mathbb{R}^n \).

We want to prove that \( f \) is Lipschitz continuous. This means there exists a constant \( L \) such that for all \( x, y \in \mathbb{R}^n \):
\[
\|f(x) - f(y)\| \leq L \|x - y\|.
\]

### 2. Use the Mean Value Theorem
The Mean Value Theorem for vector-valued functions states that for any \( x, y \in \mathbb{R}^n \), there exists a point \( c \) on the line segment between \( x \) and \( y \) such that:
\[
f(x) - f(y) = Df(c) \cdot (x - y),
\]
where \( Df(c) \) is the derivative of \( f \) at the point \( c \).

### 3. Take Norms on Both Sides
Taking the norm on both sides of the above equation, we get:
\[
\|f(x) - f(y)\| = \|Df(c) \cdot (x - y)\|.
\]

### 4. Apply the Boundedness of the Derivative
We know that \( \|Df(c)\| \leq M \). Therefore:
\[
\|f(x) - f(y)\| \leq \|Df(c)\| \cdot \|x - y\| \leq M \|x - y\|.
\]

### 5. Conclusion
The inequality \( \|f(x) - f(y)\| \leq M \|x - y\| \) holds for all \( x, y \in \mathbb{R}^n \). This shows that \( f \) is Lipschitz continuous with Lipschitz constant \( L = M \).

Hence, we have proven that \( f \) is Lipschitz continuous.

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Do you want details or have any questions about this proof? 

Here are 5 related questions you might be interested in:

1. What is the significance of a function being Lipschitz continuous?
2. How does Lipschitz continuity differ from uniform continuity?
3. Can a function be Lipschitz continuous if its derivative is unbounded?
4. What are some examples of Lipschitz continuous functions?
5. How does the Mean Value Theorem apply to functions that are not differentiable everywhere?

**Tip:** Remember, a function that is Lipschitz continuous is always uniformly continuous, but the converse is not necessarily true.

Learn how to prove Lipschitz continuity of a continuously differentiable function given the bounded norm of its derivative. This problem involves advanced concepts such as the Mean Value Theorem for vector-valued functions. Suitable for advanced undergraduate students and researchers.

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