I have to prove that $g$ is continous but my TA has told me my argumentation is not correct. Can you guys please explain where it goes wrong? I know the following:
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be continous and let $a,b \in \mathbb{R}^n$. The function $g: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $$g(t) = f(ta+(1-t)b)$$
I have done the following. Let $(t_k)_{k=1}^\infty \rightarrow r$ for $k \rightarrow \infty$. As $f$ is continous we have \begin{align}\lim_{k \rightarrow \infty} f(t^k) &= f(\lim_{k \rightarrow \infty} t^k)\\ &= f(\lim_{k \rightarrow \infty} (t^k+(1-t^k)b)) = f(r)\end{align} but he has told me that this does not make any sense? I have used the above to say that \begin{align}\lim_{k \rightarrow \infty} g(t^k) &= \lim_{k \rightarrow \infty} f(t^ka+(1-t^k)b)\\ &= f(\lim_{k \rightarrow \infty} (t^ka+(1-t^k)b))\\ &= f(ra+(1-t)b)\end{align} and we are done as $g(t^k) \rightarrow g(r)$ and therefore is continous. Is this not correct? And if so what should I be doing different.
$f(lim((t_k)$ is not equal to $f(lim(t_ka+(1-t_k)b)$ you have to write $lim(g(t_k))=limf(t_ka+(1-t_k)b)=f(lim(t_ka+(1-t_k)b)=f(ra+(1-r)b)=g(r)$.