Exercise 7.5.5

Question

The Fundamental Theorem of Calculus can be used to supply a shorter argument for Theorem 6.3.1 under the additional assumption that the sequence of derivatives is continuous.

Assume $f_{n} \rightarrow f$ pointwise and $f_{n}^{\prime} \rightarrow g$ uniformly on $[a, b]$. Assuming each $f_{n}^{\prime}$ is continuous, we can apply Theorem 7.5.1 (i) to get
$$
\int_{a}^{x} f_{n}^{\prime}=f_{n}(x)-f_{n}(a)
$$
for all $x \in[a, b]$. Show that $g(x)=f^{\prime}(x)$.

Ulisse Mini · Accepted Answer

Since $f'_n \to g$ uniformly, $g$ is continuous. The Integrable Limit Theorem tells us $\int^x_a f'_n \to \int^x_a g = f(x) - f(a)$, and Theorem 7.5.1 (ii) tells us $G(x) = f(x) - f(a)$ satisfies $$G'(x) = f'(x) = g(x)$$
(since $f(a)$ is constant).

Exercise 7.5.5

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Exercise 7.5.5

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