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I came across the following problem in a prelim question paper. The question as stated seems meaningless to me, I am adding the picture so as to avoid any error from my end. Question

My case with the above question is that $f$ is given to be defined on $(0,1)$ so $f(x)$ makes sense only if $x\in (0,1)$. Although, it doesn’t tell s anything but if we assume therefore that the range is contained only in $(0,1)$. But then it would mean that a point $x$ is taken only $0$ times which is again a nonsense.

I am not sure if I am missing something or the question is really wrong. If it is wrong, what can be “the nearest correct” version of it? Meaning if we take the function to be defined from $\mathbb{R}$ to $(0,\infty)$?

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    $\begingroup$ I agree with you that the part that says, “for every $x \in \mathbb{R}$, the preimage of $x$ has at most $(f(x))^2$ elements” doesn’t make sense, because that’s only defined for $x \in (0,1)$. I don’t know what the “nearest correct” version is. $\endgroup$ – Joe Aug 22 '19 at 9:51
  • $\begingroup$ If the issue raised by Joe is not solved, the question is meaningless as it stands. A shame, since it sounds very interesting. $\endgroup$ – астон вілла олоф мэллбэрг Aug 22 '19 at 10:56
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    $\begingroup$ It seems like a typo, the correct question is likely that for every $x\in \Bbb R$ the set $f^{-1}(x)$ has at most $x^2$ elements. The typo likely arises from a confusion, as conventionally $x$ is used to denote points in the domain, so the statement might as well be "$f^{-1}(x)$ has at most $f(y)^2$ many elements for $y\in f^{-1}(x)$". $\endgroup$ – s.harp Aug 22 '19 at 11:05
  • $\begingroup$ @s.harp : I read this as approximating a different question. "for every $y \in \mathbb{R}$, the set $f^{-1}(y)$ has at most $(f^{-1}(y))^2$ elements" meaning that there cannot be many points in the image with small magnitude. Suppose $x$ is such that $f(x) = 9.5$, then there are at most $9.5^2 = 90.25$ points in $(0,1)$ whose image through $f$ is $9.5$. $\endgroup$ – Eric Towers Aug 23 '19 at 15:15
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    $\begingroup$ @EricTowers your first sentence doesn't make sense, but the example you give is exactly what I wrote. $\endgroup$ – s.harp Aug 23 '19 at 15:36
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As others have said, the question doesn't make sense as stated, but what we can prove is that if $f: (0,1) \rightarrow \mathbb{R}$ is continuous and $|f^{-1}(x)|\leq x^2$, then $f$ is differentiable a.e.. For that it will suffice to show that $f$ is differentiable a.e. on an interval of the form $[1/n, 1-1/n]$.

Note that $|f|$ is bounded on such an interval, say by $N\in\mathbb{N}$ so by the assumption $f$ attains any value at most $N^2$ times on $[1/n, 1-1/n]$. This implies that $f$ is of bounded variation on said interval: if $1/n=a_0<a_1<...<a_k=1-1/n$, note that by the intermediate value theorem, $f$ will for each $i$ attain all values in the interval $[f(a_{i-1},f(a_i)]$. Hence the length of all of these intervals cannot exceed $2N^3$. Indeed, if it did, then since each $[f(a_{i-1}),f(a_i)]$ is contained in the interval $[-N, N]$ (of length $2N$) the pigeonhole principle implies that at least one $x\in[-N,N]$ must occur in $N^2+1$ of the $[f(a_{i-1}),f(a_i)]$ intervals-this is a contradiction. Hence $\sum\limits_{i=1}^k|f(a_i)-f(a_{i-1})|\leq 2N^3$ so since $1/n=a_0<a_1<...<a_k=1-1/n$ was an arbitrary partition of $[1/n,1-1/n]$, $f$ is of bounded variation thereon. Hence it is a.e. differentiable on $[1/n,1-1/n]$.

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