Suppose $y=f(x)$ is a continuous function and $f(x)=f(x')$ with $x≠x'$. Can we always find a sub-interval of the interval $[x, x']$ where $f$ is a simple hump or trough? By a simple hump, I mean a curve that rises monotonically from a certain height $y=k$, reaches a maximum, and then falls monotonically back to $y=k$. A simple trough is the inverse of that.
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No, we can't necessarily do that. Take, for instance, the Weierstrass function, whose graph is a fractal, going up and down infinitely many times on any interval.
Note that if your function is indeed a bumb, meaning that it is first increasing then decreasing, then it is necessarily differentiable almost everywhere.
Arthur
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Between the continuous functions and the differentiable functions, there is a whole world.
If we denote by $\mathscr C^0([a,b])$ and $\mathscr C^1([a,b])$ the sets of all continuous functions and differentiable (with continuity) function on $[a,b]$, we have that $$ \mathscr C^1([a,b])\subset\mathscr C^0([a,b]) $$
Now, between them you can find the so called $\alpha$-Holder functions; we denote this set with $\mathscr C^{\alpha}([a,b])$, where $0<\alpha\le1$. This set is characterized by all the functions $f:[a,b]\to\Bbb R$ such that $$ \sup_{s,t\in[a,b]\\s\neq t}\frac{|f(t)-f(s)|}{|t-s|^{\alpha}}<+\infty $$ and it is easy to prove that, for every $0<\alpha<\beta\le1$ $$ \mathscr C^1([a,b])\subset\mathscr C^{\beta}([a,b])\subset\mathscr C^{\alpha}([a,b])\subset\mathscr C^0([a,b]) $$ The simplest function belonging to $\subset\mathscr C^{\alpha}([a,b])$ is $x\mapsto\sqrt{x-a}$, even if, this function doesn't catch the irregular nature that other Holder functions show.
A more significative example is given by the paths of fractional Brownian motion of Hurst index $0<H<1$.
Joe
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