# 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 ${{\mathbb{R^n}}}$?

• Keep going with your trick of breaking things up. The sum of continuous functions is continuous, and the product of continuous functions is continuous, and constant functions are continuous. Feb 3, 2018 at 21:32
• $a,b\in \mathbb R$ makes no sense.
– zhw.
Feb 3, 2018 at 23:43

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$$|t_2-t_1|<\delta\implies\bigl\|h(t_2)-h(t_1)\bigr\|<\varepsilon.$$

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 $\mathbb{R} \to \mathbb{R}^n$.

$h: \mathbb{R} \rightarrow \mathbb{R^n}$

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

$|h(t)-h(t_0)| =|(t-t_0)| ||(a-b)||$

$\lt \delta ||a-b|| \lt \epsilon$.