Thursday, May 10, 2018

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


Answer



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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