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A function $f:\mathbb{R^+}→\mathbb{R}$ satisfies the condition $f(ab)=f(a)+f(b)$ for $a,b>0$. If $f(x)$ is continuous at $x=1$, then prove that $f$ is continuous on $\mathbb{R^+}$.

Attempt:

I have proved $f(1)=0$, $f(1/c)=-f(c),~~ c>0$.

Consider a sequence $\{x_n\}$ in $\mathbb{R^+}$ converges to $c>0$, then $$\lim_{n\to \infty}f(x_n)=\lim_{n\to \infty}f(\frac{x_n}{c}c)=\lim_{n\to \infty}f(\frac{x_n}{c})+\lim_{n\to \infty}f(c)$$

how to proceed further to use sequential criterion of continuity to prove that $f$ is continuous on $\mathbb{R^+}$.

My aim is to use sequential criterion of continuity $x_n\to c$ and $f(x_n)\to f(c)$ implies $f$ is continuous.

Please help.

Martin Sleziak
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rama_ran
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  • https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – carmichael561 Apr 17 '17 at 18:20
  • I'm not sure how to answer this (still figuring it out), but note that it is sufficient to show continuity at $1$. Also, local boundedness at $1$ is sufficient for continuity there. Suppose there exists $M > 0$ such that $|f(x)| \le M$ for $x \in (1- \delta, 1 + \delta)$. Let $r = 1 + \delta$. Then if $y > 1$ and $\log_r (y) = t$, $lf(y)| \le tM$. It follows that as $t \to 0$ (i. e. as $y \to 1$, $f(y)$ goes to zero. – Vik78 Apr 17 '17 at 19:11
  • With some effort it can proved that $f$ is differentiable also on $\mathbb{R}^{+}$. See the theorem mentioned in https://math.stackexchange.com/a/2091337/72031 – Paramanand Singh Apr 18 '17 at 09:26

2 Answers2

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Note that $f(1) = 0$. Fix $x > 0$. We have: $$f(x+h) - f(x) = f \left( 1 + \frac{h}{x} \right)$$

We have $1 + \frac{h}{x} \to 1$ as $h \to 0$, so by continuity at $1$, $f\left( 1 + \frac{h}{x} \right) \to f(1) = 0$. Therefore $f$ is continuous at $x$.

  • well. Thaks a lot. Would you please suggest me the solution using sequential criterion of continuity, as I mentioned above? – rama_ran Apr 17 '17 at 18:31
  • @rama_ran As far as your question posted here is concerned, '$;1 + \frac{h}{x} \to 1$ as $h \to 0;$' is shorthand for 'if we have a sequence $(h_n){n \in \mathbb{N}}$ sucht that $lim_n h_n = 0$, then $lim_n (1 + \frac{h_n}{x} \to 1);$' (in $\mathbb{R+}$ and $\mathbb{R}$, respectively). – polynomial_donut Apr 17 '17 at 20:27
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Using your desired approach, we have by conditinuity of scalar multiplication that $x_n/c\to c/c=1$. So, as $f$ is continuous at $1$ by hypothesis, we have, as you wrote, $$\lim_{n\to \infty}f(x_n)=\lim_{n\to \infty}f\left(\frac{x_n}{c}c\right)=\lim_{n\to \infty}f\left(\frac{x_n}{c}\right)+\lim_{n\to \infty}f(c)=f(1)+f(c)=f(c)$$ (as $f (1)=f(1\cdot 1)=f(1)+f(1)$ implies $f(1)=0$) and so $f$ is continuous at $c$ as for any sequence $\{x_n\}$ converging to $c$, we have that $f(x_n)\to f(c)$.

Josué Tonelli-Cueto
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