Let $f:\mathbb{R} \rightarrow \mathbb{R}, \\ f(0)=-1, \\ f(x+y) \leq -f(x)f(y).$

Show that $f \text{ continuous in } \mathbb{R} \iff f \text{ continuous in } 0$.

$\Rightarrow$ is trivial, as $0 \in \mathbb{R}$.

$\Leftarrow$ is pretty hard for me.

You could begin with saying that for all $\varepsilon > 0$, there is a $\delta > 0$ such that $|f(x)-f(0)|=|f(x)+1| < \varepsilon$ for all $|x| < \delta$.

But how do you go on? Or is there simply a better solution?

Thanks in advance!

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    $\begingroup$ Presumably you meant "$...=|f(x)+1|<\varepsilon$" when $|x|<\delta.$ $\endgroup$ – Thomas Andrews Dec 11 '19 at 19:37
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    $\begingroup$ If you let $g(x)=-f(x)$ then $g(0)=1$ and $g(x+y)\geq g(x)g(y).$ Not sure how that helps, but it seems like a slightly easier problem, and $g$ is continuous at $0$ iff $f$ is, and $g$ is continuous on all of $\mathbb R$ when $f$ is. $\endgroup$ – Thomas Andrews Dec 11 '19 at 19:41
  • $\begingroup$ Letting $g(x)=-f(x)$ you can show that $g(x)>0$ for all $x$ thus you can define $h(x)=\log g(x).$ Then $h$ is continuous at $0,$ $h(x+y)\geq h(x)+h(y)$ and $h(0)=0.$ We have that $h$ is continuous on $\mathbb R$ iff $g$ is. Not sure if that helps. $\endgroup$ – Thomas Andrews Dec 11 '19 at 20:12
  • $\begingroup$ I'll try to think about it that way, thank you. $\endgroup$ – marymk Dec 11 '19 at 20:21

Following the idea from Thomas's comment, define $g = -f$ and verify that $g(0) = 1$ and $g(x+y)\geqslant g(x)g(y)$.

Any $x\in \Bbb R$ can be written as $n\epsilon$ for some $\epsilon$ small enough and suitable $n$. Then, because $g(0) = 1 > 0$ and by continuity of $g$ at $0$ we'll have that $g(\epsilon)>0$. Therefore

$$g(n\epsilon) \geqslant g((n-1)\epsilon)g(\epsilon) \geqslant g((n-2)\epsilon)g(\epsilon)^2 \geqslant \dots \geqslant g(\epsilon)^n > 0.$$

In other words, $g>0$.

Now, write

$$g(c) = g(c+h-h) \geqslant g(c+h)g(-h) \geqslant g(c)g(h)g(-h)$$

so that

$$g(c)g(h)\leqslant g(c+h) \leqslant g(c)/g(-h).$$

Notice that $g>0$ so the division by $g(-h)$ and inequality signs are okay.
Additionally, $1=g(0)\geqslant g(h)g(-h)$ so that $g(c)g(h) \leqslant g(c)/g(-h)$ indeed.

Now, from $g(0) = 1$, the continuity of $g$ at $0$, and the squeeze theorem, for any fixed $c$ we get $\lim_{h\to 0}g(c+h) = g(c)$, which implies the continuity of $g$ at $c$ as desired.

  • $\begingroup$ Nicely done! Comments are impermanent, so you should define $g(x)$ explicitly in your answer. $\endgroup$ – Thomas Andrews Dec 11 '19 at 20:26
  • $\begingroup$ Thank you! You are right, I will edit this in. $\endgroup$ – Fimpellizieri Dec 11 '19 at 20:41
  • $\begingroup$ Can you explain why given $x \in \mathbb{R}$ there is $\epsilon$ and $n$ such that $g(n\epsilon) = x$? Did you perhaps mean that $x = n\epsilon$, for suitable $n$ and $\epsilon$? $\endgroup$ – Nicholas Roberts Dec 12 '19 at 2:13
  • $\begingroup$ Yes, you are right. That would obviously be wrong given the conclusion that $g>0$ haha. $\endgroup$ – Fimpellizieri Dec 12 '19 at 13:05

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