Consider a function $ \gamma(t) = (f_1(t), f_2(t), ... , f_k(t))$ for $t \in [a,b]$. So, $f_i$ are real-valued functions on $[a,b]$. Then $\gamma$ is continuous, i.e. a path, if and only if each $f_i$ is continuous.
Show that if the functions $f_i$ are uniformly continuous, then so is $\gamma$.
Can anyone give me a tip on where to start for this problem?
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Let $t_0\in[a,b]$ and let $\epsilon>0$. For each $f_i$ there exists $\delta_i$ such that if $|t-t_0|<\delta_i)$ then $|f_i(t)-f_i(t_0)|<\epsilon$.
Let $\delta=\min\{\delta_i:1\le i\le k\}$. Then if $|t-t_0|<\delta$, each $|f_i(t)-f_i(t_0)|<\epsilon$.
Using the euclidian metric, $|\gamma(t)-\gamma(t_0)|\le(k\epsilon^2)^{1/2}=\sqrt{k}\epsilon$, so $\gamma$ is continuous at $t_0$.
For uniform continuity, all you have to do is find $\delta_i$ that works for all $t\in[a,b]$. I'll let you fill in the details.
How your proof should look: let $\epsilon>0$, $t \in [a,b]$. For each $f_j$ we can find a $\delta_j>0$ such that for $s \in [a,b]$ such that for $|t - s| < \delta_j$ we have $|f(t) - f(s)|<\epsilon$. Set $\delta = \min \{\delta_j\}$.
I'm not sure what norm you're using. You should pick "a nice $\epsilon$" so that when $s$ and $t$ are close enough you'll have that the distance between $\gamma(t)$ and $\gamma(s)$ is small. The proof for uniform continuity should follow a nearly identical argument.