2 replaced http://math.stackexchange.com/ with https://math.stackexchange.com/ edited Apr 13 '17 at 12:20 The question is almost a duplicate of http://math.stackexchange.com/q/71121/Space of bounded continuous functions is complete but it's easier because continuous functions on $$[a,b]$$ are automatically bounded. So, I repeat the proof by Matt N.the proof by Matt N. with slight modifications. Given a Cauchy sequence $$(f_n)$$ in $$C([a,b];\mathbb R^n)$$, we first show that the sequence has a pointwise limit. For this we note that because $$f_n$$ is Cauchy with respect to the supremum norm, it follows that $$f_n(x)$$ is a Cauchy sequence in $$\mathbb{R}^n$$ for any $$x$$ in $$[a,b]$$. But $$\mathbb{R}^n$$ is complete and hence the limit $$\lim_{n \to \infty} f_n (x)$$ exists in $$\mathbb{R}^n$$. Let $$f(x)$$ denote this limit. Now we want to show that $$f_n$$ converges to $$f$$ uniformly, that is $$\sup_{[a,b]} \| f - f_n |\ \to 0$$. Given $$\varepsilon > 0$$, we have $$N$$ such that for $$n,m \geq N$$, $$\|f_n(x) - f_m(x) \| < \frac{\varepsilon}{2}$$, again because $$f_n$$ is Cauchy. Passing to the limit $$m\to\infty$$ we get $$\|f_n(x) - f (x) \| \le \frac{\varepsilon}{2}<\epsilon$$. Finally, now that we have convergence in norm, we can apply the uniform limit theorem (proof here) to get that $$f$$ is continuous and hence $$f$$ is in $$C([a,b];\mathbb R^n)$$. The question is almost a duplicate of http://math.stackexchange.com/q/71121/ but it's easier because continuous functions on $$[a,b]$$ are automatically bounded. So, I repeat the proof by Matt N. with slight modifications. Given a Cauchy sequence $$(f_n)$$ in $$C([a,b];\mathbb R^n)$$, we first show that the sequence has a pointwise limit. For this we note that because $$f_n$$ is Cauchy with respect to the supremum norm, it follows that $$f_n(x)$$ is a Cauchy sequence in $$\mathbb{R}^n$$ for any $$x$$ in $$[a,b]$$. But $$\mathbb{R}^n$$ is complete and hence the limit $$\lim_{n \to \infty} f_n (x)$$ exists in $$\mathbb{R}^n$$. Let $$f(x)$$ denote this limit. Now we want to show that $$f_n$$ converges to $$f$$ uniformly, that is $$\sup_{[a,b]} \| f - f_n |\ \to 0$$. Given $$\varepsilon > 0$$, we have $$N$$ such that for $$n,m \geq N$$, $$\|f_n(x) - f_m(x) \| < \frac{\varepsilon}{2}$$, again because $$f_n$$ is Cauchy. Passing to the limit $$m\to\infty$$ we get $$\|f_n(x) - f (x) \| \le \frac{\varepsilon}{2}<\epsilon$$. Finally, now that we have convergence in norm, we can apply the uniform limit theorem (proof here) to get that $$f$$ is continuous and hence $$f$$ is in $$C([a,b];\mathbb R^n)$$. The question is almost a duplicate of Space of bounded continuous functions is complete but it's easier because continuous functions on $$[a,b]$$ are automatically bounded. So, I repeat the proof by Matt N. with slight modifications. Given a Cauchy sequence $$(f_n)$$ in $$C([a,b];\mathbb R^n)$$, we first show that the sequence has a pointwise limit. For this we note that because $$f_n$$ is Cauchy with respect to the supremum norm, it follows that $$f_n(x)$$ is a Cauchy sequence in $$\mathbb{R}^n$$ for any $$x$$ in $$[a,b]$$. But $$\mathbb{R}^n$$ is complete and hence the limit $$\lim_{n \to \infty} f_n (x)$$ exists in $$\mathbb{R}^n$$. Let $$f(x)$$ denote this limit. Now we want to show that $$f_n$$ converges to $$f$$ uniformly, that is $$\sup_{[a,b]} \| f - f_n |\ \to 0$$. Given $$\varepsilon > 0$$, we have $$N$$ such that for $$n,m \geq N$$, $$\|f_n(x) - f_m(x) \| < \frac{\varepsilon}{2}$$, again because $$f_n$$ is Cauchy. Passing to the limit $$m\to\infty$$ we get $$\|f_n(x) - f (x) \| \le \frac{\varepsilon}{2}<\epsilon$$. Finally, now that we have convergence in norm, we can apply the uniform limit theorem (proof here) to get that $$f$$ is continuous and hence $$f$$ is in $$C([a,b];\mathbb R^n)$$. 1 answered Jun 22 '13 at 0:43 ˈjuː.zɚ79365 2,42411 gold badge2222 silver badges119119 bronze badges The question is almost a duplicate of http://math.stackexchange.com/q/71121/ but it's easier because continuous functions on $$[a,b]$$ are automatically bounded. So, I repeat the proof by Matt N. with slight modifications. Given a Cauchy sequence $$(f_n)$$ in $$C([a,b];\mathbb R^n)$$, we first show that the sequence has a pointwise limit. For this we note that because $$f_n$$ is Cauchy with respect to the supremum norm, it follows that $$f_n(x)$$ is a Cauchy sequence in $$\mathbb{R}^n$$ for any $$x$$ in $$[a,b]$$. But $$\mathbb{R}^n$$ is complete and hence the limit $$\lim_{n \to \infty} f_n (x)$$ exists in $$\mathbb{R}^n$$. Let $$f(x)$$ denote this limit. Now we want to show that $$f_n$$ converges to $$f$$ uniformly, that is $$\sup_{[a,b]} \| f - f_n |\ \to 0$$. Given $$\varepsilon > 0$$, we have $$N$$ such that for $$n,m \geq N$$, $$\|f_n(x) - f_m(x) \| < \frac{\varepsilon}{2}$$, again because $$f_n$$ is Cauchy. Passing to the limit $$m\to\infty$$ we get $$\|f_n(x) - f (x) \| \le \frac{\varepsilon}{2}<\epsilon$$. Finally, now that we have convergence in norm, we can apply the uniform limit theorem (proof here) to get that $$f$$ is continuous and hence $$f$$ is in $$C([a,b];\mathbb R^n)$$. Post Made Community Wiki by ˈjuː.zɚ79365 occurred Jun 22 '13 at 0:43