# Let $f(0) = -1, f(x+y) \leq -f(x)f(y)$, show that $f \text{ continuous in } \mathbb{R} \iff f \text{ continuous in } 0$

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?

• Presumably you meant "$...=|f(x)+1|<\varepsilon$" when $|x|<\delta.$ – Thomas Andrews Dec 11 '19 at 19:37
• 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. – Thomas Andrews Dec 11 '19 at 19:41
• 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. – Thomas Andrews Dec 11 '19 at 20:12
• I'll try to think about it that way, thank you. – 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.

• Nicely done! Comments are impermanent, so you should define $g(x)$ explicitly in your answer. – Thomas Andrews Dec 11 '19 at 20:26
• Thank you! You are right, I will edit this in. – Fimpellizieri Dec 11 '19 at 20:41
• 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 $\epsilon$? – Nicholas Roberts Dec 12 '19 at 2:13
• Yes, you are right. That would obviously be wrong given the conclusion that $g>0$ haha. – Fimpellizieri Dec 12 '19 at 13:05