What conditions should be imposed on a continuous function defined in $\mathbb{R}$ in order for $\rho(x,y) = |f(x)-f(y)|$ to be a metric and $\mathbb{R}$ to be a complete space.
In order for it to be a metric, we need to have $\rho = 0$ iff $x=y$ so $f$ needs to be injective. I think the rest of the things necessary for a metric are all fulfilled because we have an absolute value, but I'm not 100% sure about triangle inequality.
In order for $\mathbb{R}$ to be complete, we need that every Cauchy sequence in $\mathbb{R}$ converges to a point in $\mathbb{R}$. I can't figure out what the condition here is. f being bounded maybe? Any help is appreciated
Triangle inequality is always satisfied, as $$\rho(x, y) + \rho(y, z) = |f(x) - f(y)| + |f(y) - f(z)| \ge |f(x) - f(z)| = \rho(x, z).$$ To get completeness, what you need is a closed range. Let's suppose the range of $f$ is closed, and $(x_n)$ is Cauchy under $\rho$. Then $$0 = \lim_{n, m \to \infty} \rho(x_n, x_m) = \lim_{n, m \to \infty} |f(x_n) - f(x_m)|.$$ It follows that $(f(x_n))$ is Cauchy under the Euclidean metric, and converges to some $y$. But this is a sequence in the range of $f$, which is closed, hence $y = f(x)$ for some $x$. It's easy to show therefore that $x_n \to x$ under $\rho$.
Conversely, if $\rho$ is complete, and $f(x_n) \to y$ in the Euclidean metric, then $(f(x_n))$ is Cauchy under the Euclidean metric. Thus $(x_n)$ is Cauchy under $\rho$, which means $x_n \to x$ for some $x$. It follows therefore that $f(x_n) \to f(x)$, hence the range of $f$ is closed.