Since $f(x,t)$ is uniformly continuous, for any $\epsilon > 0$, we can find $\delta$ so that $\left|f(x,t) - f(x_0, t)\right| < \frac{\epsilon}{d-c}$ whenever $\|(x,t) - (x_0, t) \| = \left|x - x_0\right| < \delta$. Then when $\left|x - x_0\right| < \delta$, \[\left|F(x) - F(x_0)\right| \leq \int_c^d \left|f(x, t) - f(x_0, t)\right| dt < \int_c^d \frac{\epsilon}{d-c} dt = \epsilon \] showing $F(x)$ is continuous.

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Understanding Analysis

Exercise 8.4.13

Prove Theorem 8.4.5.

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Since $f (x, t)$ is uniformly continuous, for any $𝜖 > 0$ , we can find $δ$ so that $| f (x, t) - f (x_{0}, t) | < \frac{𝜖}{d - c}$ whenever $∥ (x, t) - (x_{0}, t) ∥ = | x - x_{0} | < δ$ . Then when $| x - x_{0} | < δ$ ,

| F (x) - F (x_{0}) | \leq \int_{c}^{d} | f (x, t) - f (x_{0}, t) | dt < \int_{c}^{d} \frac{𝜖}{d - c} dt = 𝜖

showing $F (x)$ is continuous.

ulissemini

2022-01-27 00:00

Exercise 8.4.13

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