I'm struggling with an assignment on function continuity. I'm asked to show two things:

Part A) Let $f:R\rightarrow R$ be a function where $f(x+y) = f(x)f(y)$ for all $x,y \in R$

I'm asked to prove that if $f$ is continuous at $x_0=0$ then $f$ is continuous at all points $x_0 \in R$

I'm rather at a loss here. For the first part (A), we see that

$$||f(x_0)-f(x)||=||f(x_0 + (x-x_0))-f(x_0)||=||f(x_0)f(x-x_0)-f(x_0)||$$

Continuity at $x_0$ implies that for any $\epsilon>0$ there exists a $\delta>0$ such that when

$$||x-x_0||<\delta \rightarrow ||f(x)-f(x_0)||<\epsilon$$

I know this holds for $x_0=0$ but I don't see how this helps to prove general continuity?

Part B) Let $g:R \rightarrow R$ with $g(xy)=g(x)+g(y)$. And I'm similarly to prove that if $g$ is contious at $x_0=1$ then it is continous for all $x_0>0$

I suspect this second part will be elucidated, if I manage to understand the first, so I'll let this rest for now. Noting only that $$||g(x)-g(x_0)||=||g\big(x_0\frac{x}{x_0}\big)-g(x_0)||=||g(x_0)+g(\frac{x}{x_0})-g(x_0)||=||g(\frac{x}{x_0})||$$

But again I'm not sure how this leads to continuity at all $x_0>0$.


You are close. Note that for both questions, $f \equiv 0$ and $g \equiv 0$ are possible solutions, in which then they are clearly continuous. If $f \not\equiv 0$, then $f(0) = 1$. Fix $\epsilon > 0$, and choose $\delta$ such that $|x| < \delta \Rightarrow |f(x) - 1| < \frac{\epsilon}{|f(x_0)|}$. We then have $|x - x_0| < \delta \Rightarrow |f(x - x_0) - 1| < \frac{\epsilon}{|f(x_0)|}$. Following your working:

\begin{align*} |f(x) - f(x_0)| = |f(x_0)f(x - x_0) - f(x_0)| = |f(x_0)||f(x - x_0) - 1| < \epsilon \end{align*}

For part b), you have $g(1) = 0$ if $g \not\equiv 0$. The working is very similar, so you can try it out yourself.

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  • $\begingroup$ I can follow the part of $|x| < \delta \Rightarrow |f(x) - 1| < \frac{\epsilon}{|f(x_0)|}$, as this is simply continuity at $x_0=0$ with a rewritten norm. However, I don't see how you get "$|x - x_0| < \delta \Rightarrow |f(x - x_0) - 1| < \frac{\epsilon}{|f(x_0)|}$". The relation holds for $x_0=0$ that $|f(x)-1|<\frac{\epsilon}{|f(x_0)}$, but how does that translate to holding for any $x_0$? might not $|f(x-x_0)-1|$ be greater than $|f(x)-1|$? $\endgroup$ – Woodenplank Nov 21 '19 at 18:18
  • $\begingroup$ For that, I am simply substituting $x$ with $x - x_0$, so it does not matter which $x_0$ it is to start with. $\endgroup$ – Clement Yung Nov 22 '19 at 1:46
  • $\begingroup$ But the form $|f(x) - 1| < \frac{\epsilon}{|f(x_0)|}$ was derived for the specific case of $x_0=0$ where I knew, from the assignment, that $f$ was continuous. Maybe I'm just too hazy on the theory/definition, but I don't see how something derived for a specific $x_0=0$ translates to working for any $x-x_0$? $\endgroup$ – Woodenplank Nov 22 '19 at 14:14
  • $\begingroup$ You seem to be distracted/confused by $x_0$ being there, when in fact it does not matter. Try seeing it from this angle: Substitute $y := x - x_0$. Then, as stated in the earlier part of the comment, you agree that $|y| < \delta \Rightarrow |f(y) - 1| < \frac{\epsilon}{|f(x_0)|}$. Now substitute back $y$ with $x - x_0$, and my claim follows. $\endgroup$ – Clement Yung Nov 23 '19 at 12:16
  • 1
    $\begingroup$ More precisely, for any $x_0$ that we chose, if $|x - x_0| < \delta$, then $|f(x - x_0) - 1| < \frac{\epsilon}{|f(x_0)|}$. What is important is that it is the same $x - x_0$ in the $\delta$ range and in $f$. For instance, if $x = \frac{\delta}{2}$, then $|f(x) - 1| < \frac{\epsilon}{|f(x_0)|}$ as $\frac{\delta}{2} < \delta$. Similarly, if $x - x_0 = \frac{\delta}{2}$, then $|f(x - x_0) - 1| < \frac{\epsilon}{|f(x_0)|}$ as $\frac{\delta}{2} < \delta$. $\endgroup$ – Clement Yung Nov 23 '19 at 13:24

By putting $x=0=y$ we have $$f(x+y)=f(x)f(y) \implies f(0)=f^2(0) \implies f(0)=0 ~or~ 1.$$ By putting $y=0$, we get $f(x)=f(x) f(0)= f(x)$, thus $f(0)=1$.

Next, the right limit of $f(x)$ as $h\rightarrow 0$ $$f(0^+)=\lim_{h \rightarrow 0} f(0+h)=\lim_{h \rightarrow 0} f(0)f(h)=\lim_{h \rightarrow 0} f(h) ~~~~~~~(1)$$ Smilarly the left limit $$f(0^-)=\lim_{h \rightarrow 0} f(0-h)=\lim_{h \rightarrow 0} f(0)f(-h)=\lim_{h \rightarrow 0} f(-h) ~~~~~~~(2)$$ Given that $f(x)$ is cntinuous at $x=0$, then $$\lim_{h \rightarrow 0} f(h) = \lim_{h \rightarrow 0} f(-h)=f(0)=1~~~~~~~(3)$$ Let us show the continuity at any real number $x=a$ $$f(a^+)=\lim_{h \rightarrow 0} f(a+h) = f(a)\lim_{h \rightarrow 0} f(h)=f(a)~~~from~~(3)$$ Dimilarly the left limit $$f(a^-)=\lim_{h \rightarrow 0} f(a-h) = f(a)\lim_{h \rightarrow 0} f(-h)=f(a)~~~from~~(3)$$ This shows that the function is continuous at any real non zero value of $x$ provided iy is continuous at $x=0$.

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  • $\begingroup$ I appreciate the answer, which demonstrates the issue very well. Sadly this particular assignment belongs to a section of our notebook that comes before "limits of functions" - in other words, I am expected to solve it without using knowledge of limits --- Sadly. But I appreciate it nonetheless. It's a rather elegant solution. $\endgroup$ – Woodenplank Nov 21 '19 at 18:11

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