If we have a continous function $f:R \to R$ with the property that $f(x)\le x$ for every $x$, does the followng hold: $|f(x)-f(y)| \le |x-y|$ for every $x,y$ ? Visually, for me, it seems to be true, but I don't know how to prove it.
EDIT: This was part of a bigger problem. In the proof I arrived at the property $f(x)\le x$ and I thought It was sufficient for the conclusion. I will post the whole problem as well.
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Non-expansiveness would mean the derivative would be bounded (as the gradient of the secants would also be bounded). But, it's not difficult to think of such a function where the derivative is unbounded. For example, $$f(x) = \begin{cases}x \sin(e^{x^{-1}}) & \text{if }x \neq 0 \\ 0 & \text{if } x = 0\end{cases}.$$
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The statement $|f(x)-f(y)|<|x-y|$ means geometrically that your function never increases or decreases faster than $g(x)=x$ (on average). This cannot be ensured by just bounding it from above in this way.
Example. Take $f(x)=-x^2-10$ which in unbounded. It decreases faster and faster for larger values of $x$. E.g.
$$|f(10)-f(11)|=|100-121|=21\not\le 1=|10-11|.$$
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f(x) $\le$ x is the region below the y=x line in the Cartesian plane.
|f(x) - f(y)| is basically the distance between the two ordinates corresponding to the abscissas x and y.
Since the only condition f(x) has to satisfy is being continuous and $\le$ x, there can be infinite number of functions which DO NOT satisfy the given condition.
Example: y= -|x|
There must be some other required condition like f is differentiable for the question to be valid.
This is not true. What about $f(x) = -|2x|$? And even if you change the requirement to $|f(x)|\leq |x|$, we have functions like $f(x) = x\sin(x)$, for which $|f((2n + \frac12)\pi)-f((2n-\frac12)\pi)|$ gets quite a bit larger than $\pi$ for large $n$.