Exercise 6.2.4

Question

Review Exercise 5.2.8 which includes the definition for a uniformly differentiable function. Use the results discussed in Section 2 to show that if $f$ is uniformly differentiable, then $f^{\prime}$ is continuous.

Ulisse Mini · Accepted Answer

The definition of $f$ being uniformly differentiable tells us: for every $𝜖 > 0$ there exists a $δ > 0$ such that

| \frac{f (x) - f (y)}{x - y} - f^{'} (y) | < 𝜖 whenever 0 < | x - y | < δ

We can use this to show continuity of $f^{'}$ via a triangle inequality and exploiting the symmetry in $x, y$ .

| f^{'} (x) - f^{'} (y) | < | f^{'} (x) - \frac{f (x) - f (y)}{x - y} | + | \frac{f (x) - f (y)}{x - y} - f^{'} (y) | < 𝜖

After picking $δ$ so that every $| x - y | < δ$ has

| \frac{f (x) - f (y)}{x - y} - f^{'} (y) | < 𝜖 ∕ 2

Alternative proof: Let $y_{n} = x + 1 ∕ n$ , and consider the sequence of functions

f_{n}^{'} (x) = \frac{f (x) - f (y_{n})}{x - y_{n}}

Each $f_{n}^{'} (x)$ is continuous, and uniform differentiability implies that $f_{n}^{'}$ uniformly converges to $f^{'} (x)$ ; hence $f^{'} (x)$ is continuous by Theorem 6.2.6.

Exercise 6.2.4

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Exercise 6.2.4

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