Is there a continuous function $$u\colon (0,1)\to \mathbb{R}$$ such that at every point $x\in (0,1)$ one has $$\lim\sup_{y\to x+0}\frac{u(y)-u(x)}{y-x}=+\infty?$$
In particular $u$ is not differentiable everywhere.
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No. Indeed, suppose that such a function $u$ exists. For any real $a$ and all $x\in(0,1)$, let $u_a(x):=u(x)-ax$. Then the upper right derivative of the function $u_a$ is $\ge0$ on $(0,1)$ and hence $u_a$ is nondecreasing -- see e.g. Titchmarsh, The Theory of Functions, 2nd Ed., Example (iv) in Section 11.3. So, $u_a(1/3)\le u_a(2/3)$, that is, $u(1/3)\le u(2/3)-a/3$ for all real $a$, so that $u(1/3)-u(2/3)\le-\infty$, which contradicts the condition that $u$ is real valued. $\quad\Box$
answered Nov 20 at 21:44
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$\begingroup$ As an alternative to Titchmarsh, the upper right derivative of the function $u_a$ is $>0$ on $(0,1)$ and hence $u_a$ is increasing. For, if $u_a(x)\geq u_a(y)$ for some $x<y$, then $u_a$ has a maximum at some point $c\in[x,y)$, so $\frac{u_a(z)-u_a(c)}{z-c}\leq0$ for all $z\in(c,y)$, but this contradicts that the upper right derivative at $c$ should be $>0$. $\endgroup$– mr_e_manCommented Nov 22 at 1:11
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$\begingroup$ @mr_e_man : Thank you for your comment. $\endgroup$ Commented Nov 22 at 2:31