Prove non-existance of function having infinite derivative at every point Prove there does not exist a function $f:  \mathbb{R} \to \mathbb{R}$ such that for every $x$ we have $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \infty$
I've tried to find contradiction by looking at x neighbourhood, but with no effect.
 A: Such $f$ would be strictly increasing. Let $g(x) = f(x) + (f(0) - f(1)) \cdot (x - 1)$. Then $g$ still has infinite derivative, so it should be strictly increasing, bug $g(0) = g(1)$.
To prove that if $a < b$ then $f(a) < f(b)$, note that for any $x_0$ for some $\varepsilon > 0$ we have $f(x) > f(x_0)$ if $x \in (x_0, x_0 + \varepsilon)$ and $f(x) < f(x_0)$ if $x \in (x_0 - \varepsilon, x_0)$. Choose such $\varepsilon$ for each $x \in [a, b]$. Then $\{U_{\varepsilon_x}(x) | x \in [a, b]\}$ is an open cover of $[a, b]$ - so it has some finite subcover with centers $x_1 < x_2 < \ldots < x_n$. We can assume that none of $U_{\varepsilon_k}(x_k)$ is subset of union of others (otherwise just remove it from our cover). Then for any $k = 2, \ldots, n$ there is some $y_k \in U_{\varepsilon_{k - 1}}(x_{k - 1}) \cap U_{\varepsilon_k}(x_k) \cap (x_{k - 1}, x_k)$. Also if $x \in U_{\varepsilon_k}(x_k)$ then $f(x) < f(x_k)$ if $x < x_k$ and $f(x) > f(x_k)$ if $x > x_k$. So we have $f(a) \leqslant f(x_1) < f(y_1) < f(x_2) < f(y_2) \ldots < f(x_n) \leqslant f(b)$.
A: By replacing $f$ by $f-f(0)$, we can suppose that $f(0)=0$.
Consider the compact interval $I=[0,1]$. We prove that for all $n \in \mathbb N$, we have $f(x) \ge nx$ on $I$. This is absurd as then $f(1)$ is undefined.
For the proof, fix $n\in \mathbb N$.
For any $x \in I$, and using the hypothesis of the question, it exists $\epsilon_x >0$ such that for $y \in (x-2\epsilon_x, x+2\epsilon_x)\cap I$, we have $f(y) \ge n(y-x) + f(x)$.
As $I$ is compact we can cover it with a finite number of open intervals $J_i=(x_i-\epsilon_{x_i}, x_i+\epsilon_{x_i})$. Consider $i$ such that $0 \in J_i$ . If $1 \in J_i$, we are done. If not $x_i+\epsilon_i$ belongs to $J_l$ with $l \neq i$ and $f(y) \ge ny$ on $J_i \cup J_l$. Proceeding by induction, $1$ will be reached in a finite number of operations, which concludes our proof.
