Question

Show that if \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) is differentiable at \(\mathbf{X}_{0}\) and \(\epsilon>0,\) there is a \(\delta>0\) such that $$ \left|\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)\right| \leq\left(\left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\|+\epsilon\right)\left|\mathbf{X}-\mathbf{X}_{0}\right| \quad \text { if } \quad\left|\mathbf{X}-\mathbf{X}_{0}\right|<\delta . $$ Compare this with Lemma \(6.2 .5 .\).

Short answer

Question: Show that if F is differentiable at X_0, then for any ε > 0, there exists a δ > 0 such that the following inequality holds: $$ \left|\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)\right| \leq\left(\left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\|+\epsilon\right)\left|\mathbf{X}-\mathbf{X}_{0}\right| $$ Answer: The proof involves using the definition of differentiability, properties of the derivative, and the triangle inequality. We show that for any ε > 0, there exists a δ > 0 such that the inequality holds by establishing that the given inequality holds for the given delta and showing the relationship between F, F', X, X_0. This is done by applying the triangle inequality to the expression for |F(X) - F(X_0)| and showing that the established inequality holds.

Step-by-step solution

Definition of differentiability

First, let's recall the definition of differentiability: A function F is differentiable at X_0 if there exists a linear transformation F'(X_0) such that: $$ \lim_{\mathbf{X} \to \mathbf{X}_0} \frac{\left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right|}{\left|\mathbf{X} - \mathbf{X}_0\right|} = 0 $$

Start the proof with the definition of differentiability

According to the definition above, since F is differentiable at X_0, we have: $$ \lim_{\mathbf{X} \to \mathbf{X}_0} \frac{\left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right|}{\left|\mathbf{X} - \mathbf{X}_0\right|} = 0 \\ \Rightarrow \forall \epsilon > 0 , \exists \delta > 0: \left|\mathbf{X}-\mathbf{X}_{0}\right|<\delta \Rightarrow \frac{\left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right|}{\left|\mathbf{X} - \mathbf{X}_0\right|} <\epsilon $$ Now we want to show that the inequality given in the exercise statement is satisfied for some delta > 0.

Observe the inequality F(X) - F(X_0)

We want to show the inequality: $$ \left|\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)\right| \leq\left(\left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\|+\epsilon\right)\left|\mathbf{X}-\mathbf{X}_{0}\right| $$ Observe that: $$ \left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0)\right| = \left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0) + \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right| $$

Apply the triangle inequality

Now, applying the triangle inequality to the expression for |F(X) - F(X_0)|, we obtain: $$ \left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0)\right| \leq \left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right| + \left|\mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right| $$

Establish the inequality

Now, for the given delta, we have: $$ \frac{\left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right|}{\left|\mathbf{X} - \mathbf{X}_0\right|} <\epsilon $$ By multiplying both sides by |\mathbf{X} - \mathbf{X}_0|, we get: $$ \left|\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_0) - \mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right| <\epsilon \left|\mathbf{X} - \mathbf{X}_0\right|. $$ We also know that: $$ \left|\mathbf{F}'(\mathbf{X}_0)(\mathbf{X}-\mathbf{X}_0)\right| \leq \left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\| \left|\mathbf{X}-\mathbf{X}_{0}\right| $$ So the inequality is established: $$ \left|\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)\right| \leq\left(\left\|\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\right\|+\epsilon\right)\left|\mathbf{X}-\mathbf{X}_{0}\right| $$ We have shown that for an arbitrary ε > 0, there exists a δ > 0 such that the given inequality holds. This completes the proof.

Key concepts

Linear Transformation

Linear transformation is a core concept in calculus, integral to understanding differentiability. The linear transformation associated with a function at a point is essentially the best linear approximation of the function near that point. The notation \( \mathbf{F}'(\mathbf{X}_0) \) represents this transformation at \( \mathbf{X}_0 \).

This transformation can be thought of as a matrix operation if you imagine \( \mathbf{F}: \mathbb{R}^n \rightarrow \mathbb{R}^m \). It allows us to simplify complex multidimensional changes by focusing on linear relations.

• The linear transformation tells us how a small change in the input \( \mathbf{X} \) affects the output \( \mathbf{F}(\mathbf{X}) \) near \( \mathbf{X}_0 \).
• If a function is differentiable at a point, it means we can approximate it by its linear transformation there, with the difference vanishing faster than linear as you approach that point.

This simplifies our task, as it reduces the complexity of functions to linear algebra within a localized context. Understanding this concept deeply can ease the understanding of higher-level calculus problems.

Triangle Inequality

The triangle inequality is a fundamental concept in mathematics, especially useful when working with vector norms.In its basic form, for any vectors \( \mathbf{a} \) and \( \mathbf{b} \), the triangle inequality states:\[ \lvert \mathbf{a} + \mathbf{b} \rvert \leq \lvert \mathbf{a} \rvert + \lvert \mathbf{b} \rvert \]

This principle helps in simplifying and bounding expressions involving sums of vectors. It's called the triangle inequality because it reflects the idea that the direct path between two points (along two sides of a triangle) is never longer than any indirect path (the other side of the triangle).

• When applied to differentiability proofs, it helps in breaking down the distance of complex expressions into manageable parts.
• In its role within the step-by-step solution, it is used to separate the transformation term from the remainder, making it easier to handle each part individually.

Without the triangle inequality, proofs like the one above would become much more complex and mathematical manipulations less intuitive.

Inequality Proof

Proving inequalities is a common task in mathematics and involves logically demonstrating that an inequality holds under certain conditions. In the exercise provided, we prove that the function \( \mathbf{F} \) satisfies a specific inequality under the conditions of differentiability.

The proof hinges on the interplay between differentiability and the triangle inequality. By starting from the definition of differentiability and applying logical reasoning, we establish control over the difference between \( \mathbf{F}(\mathbf{X}) \) and \( \mathbf{F}(\mathbf{X}_0) \).

• Firstly, we express this difference in terms of linear transformation \( \mathbf{F}'(\mathbf{X}_0) \).
• Then, utilizing the triangle inequality, we split the expression into a sum of manageable partitions.
• Finally, by applying the concept of a small enough \( \delta \), the difference involving \( \epsilon \) becomes negligible, ensuring the inequality holds.

Such proofs build a bridge between theoretical definitions and practical applications, making them a valuable part of studying calculus and mathematical analysis.