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!
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.