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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
source | link

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 $C([a,b];\mathbb R^n)$.

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