Real analysis: Continuous Function Let $ f: {{\mathbb{R^n}}  \rightarrow {{\mathbb{R}} }}$ be continuous and let $a$ and $b$ be points in $ {{\mathbb{R} }} $
Let the function $g: {\mathbb{R}} \rightarrow {\mathbb{R}}$  be defined as:
$$ g(t) = f(ta+(1-t)b) $$
Show that $g$ is continuous .
If I define a function $ h(t)=ta+(1-t)b$, then I have that $g(t)=f(h(t))$
I know that $f$ is continuous, so I have to prove that $h(t)$ is continuous as a compound function of two continuous function is also continuous. 
How do I prove that $h(t)$ is continuous in ${{\mathbb{R^n}}}$? 
 A: If $t_1.t_2\in\mathbb R$, then\begin{align}\bigl\|h(t_2)-h(t_1)\bigr\|&=\bigl\|t_2a+(1-t_2)b-t_1a-(1-t_1)b\bigr\|\\&=\bigl\|(t_2-t_1)a-(t_2-t_1)b\bigr\|\\&=|t_2-t_1|.\|a-b\|.\end{align}If $a=b$, $h$ is the null function and therefore ir is continuous. Otherwise, if $\varepsilon>0$ then take $\delta=\frac{\varepsilon}{\|a-b\|}$. Then$$|t_2-t_1|<\delta\implies\bigl\|h(t_2)-h(t_1)\bigr\|<\varepsilon.$$
A: Just use $\varepsilon-\delta$ argument to solve the problem.
Choose arbitrary point $t_0$ in $\mathbb{R}$.
$$
\forall \varepsilon>0 \exists \delta>0 : |t-t_0|<\delta \Rightarrow |h(t)-h(t_0)|<\varepsilon
$$
We could derive the equality, $|h(t)-h(t_0)|=|(t-t_0)(a-b)|=|t-t_0||a-b|$. Note that it is because $|\cdot|$ is norm of $\mathbb{R}^n$.
So if we took $\delta$ as $\frac{\varepsilon}{|a-b|}$, then the argument holds.
This argument is independent of choosing the point $t_0$. So, $h(t)$is continuous function $\mathbb{R} \to \mathbb{R}^n$.
A: $h: \mathbb{R} \rightarrow \mathbb{R^n}$
Denote norm in $\mathbb{R}$ by $|\cdot |$, in $\mathbb{R^n}$ by $||\cdot||.$
Let $a \not= b,$ $ a,b,$ and $t$ be  real.
Show  $h(t)$ is cont. at $t=t_0.$
Let $\epsilon \gt 0$ be given.
$|h(t)-h(t_0)|= ||(t-t_0)(a-b)|| =|t-t_0| ||a-b||$
Choose $\delta \lt \dfrac{\epsilon}{||a-b||}$;
Then $|t-t_0|\lt \delta$ implies 
$|h(t)-h(t_0)| =|(t-t_0)| ||(a-b)|| $
$\lt \delta ||a-b|| \lt \epsilon$.
