# Series continuity at one point - prove continuity at all points

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.

• 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 $|f(x)-1|$? – Woodenplank Nov 21 '19 at 18:18
• For that, I am simply substituting $x$ with $x - x_0$, so it does not matter which $x_0$ it is to start with. – Clement Yung Nov 22 '19 at 1:46
• 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 $x-x_0$? – Woodenplank Nov 22 '19 at 14:14
• 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. – Clement Yung Nov 23 '19 at 12:16
• 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 $\frac{\delta}{2} < \delta$. – 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 $$x=0$$.

• 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. – Woodenplank Nov 21 '19 at 18:11