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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ a function such that for any $b\in\mathbb{R}$ there exists a strictly decreasing sequence of real numbers $(a_n)_{n\geq1}$, with $$\lim_{n\to+\infty} a_n=b$$ such that $$|f(a_n)-f(b)|\leq a_n-b$$ a) If $f$ is continuous, prove that $|f(x)-f(y)|\leq |x-y|$ for any $x, y \in\mathbb{R}$

b)If $f$ isn't continuous, does the property from a) still hold?

The hypothesis is telling us that for any real point there exists the right limit and it is equal to the value of $f$ in that point, or in other words that $$f(x)=f(x+0)$$ for any $x$ real number, where $f(x+0)$ represents the right limit.

b) is not so hard, we can use, for example, the floor function and prove that the property doesn't hold anymore. Point a) is, however, a little bit trickier. I tried to prove that there is no $x_0$ and $y_0$ such that $$|f(x_0)-f(y_0)|>|x_0-y_0|$$ but I failed to show anything. What should be done?

Fimpellizzeri
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razvanelda
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2 Answers2

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Suppose that $f$ is continuous and that $x<y$ satisfy $|f(x) - f(y)| > |x - y|$. Consider the set

$$S=\{t \in \mathbb R\,:\,|f(x) - f(t)| > |x - t|\}.$$

Notice that $y\in S$.

Let $h(t) = |f(x) - f(t)| - |x - t|$. Then $h$ is continuous and $S = h^{-1}(0,+\infty)$, and hence $S$ is open. Let $a_n \downarrow y$ be as in the hypotheses. Can you find a contradiction?

Fimpellizzeri
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This is a common technique to try to build global statements from local statements: the/a Barrier method.

Hint: Fix $x < y$ in $\mathbb{R}$. Consider the set $E = \{t \in [x,y] : |f(t) - f(x)| \leq |t-x|\}$. What can you say about $\sup E$?

user357980
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  • An example of the Barrier method at work is: https://math.stackexchange.com/questions/791411/every-increasing-function-from-a-certain-set-to-itself-has-at-least-one-fixed-po – user357980 Mar 07 '18 at 16:39