Vtyjkyuk

Let
$$f:mathbbRto mathbbR : f(x)=begincases a^x & textif $x in mathbbQ$ \ sup f(y):y<x text and y in mathbbQ & textif $x in mathbbR setminus mathbbQ$endcases$$ where $a>0$.

Prove that $f$ is continuous on $mathbbR$.

Hint: Use $f(x+y)=f(x)f(y)$ for $x,yinBbb Q$ and $f(x)approx 1$ for $0approx xinBbb Q$.

This is essentially a justification of the definition of the symbol $a^x$ for the case when $x$ is irrational. Moreover your definition requires $ageq 1$ and for $0<a<1$ you need to replace the $sup$ in your definition by $inf$.

The case $a=1$ as $f(x) =1$ for all $xinmathbb R $. Let's take $a>1$ and we prove that $lim_xto 0f(x)=1$ so that $f$ is continuous at $1$. Let's first observe that $f$ is increasing on $mathbb R $ (it is strictly increasing but we don't need that in the proof that follows).

We start with the standard result $lim_ntoinfty a^1/n=1$. Let $epsilon >0$ be arbitrary. Then there is a positive integer $m_1$ such that $|a^1/n-1|<epsilon$ for all $ngeq m_1$ and similarly there is a positive integer $m_2$ such that $|a^-1/n-1|<epsilon$ whenever $ngeq m_2$. Now we choose $delta=1/m, m=max(m_1,m_2)$ and if $-delta<x<delta$ then $$a^-1/m=f(-delta)leq f(x) leq f(delta) =a^1/m$$ Since both $a^-1/m$ and $a^1/m$ lie in interval $(1-epsilon, 1+epsilon) $ it follows that $f(x) $ also lies in the same interval so that $|f(x) - 1|<epsilon $ and this completes the proof that $f$ is continuous at $0$.

Next step is to prove that $f(x+y) =f(x) f(y) $ and this part is not difficult but rather long. Using the functional equation and continuity at $0$ one can then prove continuity of $f$ at any point $x$.

If $a geq 1$ there is nothing to prove as $f(x)=a^x$, $forall x$ by conitnuity of $a^x$ and the fact that $mathbbQ$ is dense in $mathbbR$.
So let $a <1$. If $a < 1$ then $f(x) = infty$ for $x > 0$ and $x in mathbbR setminus mathbbQ$ and $f$ is not continuous as let $a=1/2$. $sup(1/2)^y : y < x,y in mathbbQ geq lim_n rightarrow infty (1/2)^(-n) = infty$.