How do I prove that $ t \rightarrow f(G(t),H(t)) $ is continuous?

Question

I have to prove that the function $t \rightarrow f(G(t),H(t))$ is continuous in $[0,1]$.

I know the following: Let $D \subseteq \mathbb{R}^2$ and let $f : D \rightarrow \mathbb{R}$ be a continous function. Let $ G : [0,1] \rightarrow \mathbb{R}$ and $H : [0,1] \rightarrow \mathbb{R}$ be contonius functions such that $(G(t),H(t)) \in D \ \forall \ t \in [0,1]$.

Do I start by using the definition of continuous? $\forall \epsilon > 0 \exists \delta > 0 : |x-x_0| < \delta \Rightarrow |f(x)-f(x_0)| < \epsilon$? And if so do I just pick a random $\epsilon > 0$ and find a $\delta >0$?

I am really not sure how to start. Can you help me in the right direction?

Thanks.

Do you know the sequential definition of continuity? – Kavi Rama Murthy Feb 04 '20 at 11:53 — Kavi Rama Murthy, Feb 04 '20 at 11:53
Yes. Can I use that and if so how? – Mathias Feb 04 '20 at 12:01 — Mathias, Feb 04 '20 at 12:01

score 0 · Accepted Answer · answered Feb 04 '20 at 12:05

0

If $t_n \to t$ then $G(t_n) \to G(t)$ and $H(t_n) \to H(t)$. Hence $(G(t_n),H(t_n)) \to (G(t),H(t))$ which implies $f(G(t_n),H(t_n)) \to f(G(t),H(t))$.

answered Feb 04 '20 at 12:05

Kavi Rama Murthy

311,013

So we use the fact that the function f is contonius to prove that $f(G(t),H(t))$ is contonius in $[0,1]$ ? – Mathias Feb 04 '20 at 12:42
Yes. Proof using $\epsilon -\delta$ involves uniform continuity of $f$ restricted to compact subsets. It is much more complicated the proof using sequences. – Kavi Rama Murthy Feb 04 '20 at 12:48
Thanks for the help, Kavi. – Mathias Feb 04 '20 at 12:58

How do I prove that $ t \rightarrow f(G(t),H(t)) $ is continuous?

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