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

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    $\begingroup$ 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. $\endgroup$ Feb 3, 2018 at 21:32
  • $\begingroup$ $a,b\in \mathbb R$ makes no sense. $\endgroup$
    – zhw.
    Feb 3, 2018 at 23:43

3 Answers 3

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

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

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

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)|| $

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

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