Is there a periodic function that $f(x) \ge x$ for any $x$?
It seems not because a periodic function repeats so many time so that $f(x)$ can appear at many $x$ but I don't know how to prove or disprove it.
Is there a periodic function that $f(x) \ge x$ for any $x$?
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0Sketch a the graph of function on $(0,p]$, then extend that graph periodically. What does $f(x)\geq x$ mean to the graph? – 2017-01-08
3 Answers
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Hint: Assume such a function exists and consider the values of $f(0), f(P), f(2P),...$, and $0,P, 2P,...$, where $P$ is the function's period.
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0You mean they are increasing so they will reach the value of $f(0)$? – 2017-01-08
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0Yes, eventually they have to pass its value because $f(nP)$ is a fixed number. – 2017-01-08
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Visual 'proof'. Consider any $x_0.$ Since $f(x)>x$ we have $f(x_0) \ge x_0$. Then $f(x_0)$ is above the line y=x. But since the function is periodic, as $x\rightarrow\infty$, $f(x)$ returns to the value $f(x_0)$ infinitely many times. Eventually, if you take $y$ large enough, $x >f(x_0)$ for all $x>y$. But that means if you take a point $x>y$ for which $f(x) = f(x_0)$, you will have $f(x) = f(x_0) < x.$ (This point must exist by the periodicity)
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Take any point $(a,f(a))$ on the graph of the periodic function. Then by assumption $f(a)>a$.
But the line $y=f(a)$ crosses the line $y=x$ at the point $(f(a),f(a))$. So if $b>f(a)$ it follows that $f(b) In the graphic, the point on the left is $(a,f(a))$ and the point on the right is $(b,f(b))$.