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If I am given a function $$ f: [0,1] \rightarrow \mathbb{R} $$ which is continuous, and for which $ \int_{0}^{1} f(x) \, dx = 1$ , how do I prove the existence of $a,b\in (0,1)$ with $a<b$ such that $f(a)f(b)=1$? Also, how could I prove that there are infinitely many such pairs?

I have tried various things, such as working with the integral and trying to find some form of it to which I could apply MVT for definite integrals on an interval $[0,1/2]$, so that I could get an '$a$' in said interval and $b=1-a$, though I've had no success yet. I'm thinking the solution wouldn't even use said theorem, as the values HAVE to be inside $[0,1]$ and NOT $0$ or $1$.

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    $\begingroup$ Possible duplicate: math.stackexchange.com/questions/4877331/… $\endgroup$
    – sudeep5221
    Commented Mar 8 at 16:55
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    $\begingroup$ Hint: Prove that either $f=1$ or there exist $a,b\in (0,1)$ such that $f(a)<1<f(b)$. Then use continuity to get your points. $\endgroup$
    – Severin Schraven
    Commented Mar 8 at 17:15
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    $\begingroup$ What does the image of $f$ (that is, the range of values it takes on) look like? $\endgroup$
    – Brian Tung
    Commented Mar 8 at 17:16
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    $\begingroup$ @RemWheel: It's a hint. ;-) There is more that you can say about the image or range of $f$ (as opposed to its codomain, the reals) than simply that they are all real values. Could the image be, for example, $\{0.5, 1.5\}$? (I.e., not the interval between $0.5$ and $1.5$, but just those two specific values.) Could the image be the interval $[0, 0.75)$? And so forth. $\endgroup$
    – Brian Tung
    Commented Mar 8 at 17:34
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    $\begingroup$ @RemWheel: Yes, now you're thinking the right way! $\endgroup$
    – Brian Tung
    Commented Mar 8 at 17:41

1 Answer 1

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One has to look at the function $g(x,y) = f(x)\cdot f(y)$ and integrate it over $[0,1]\times[0,1]$. Clearly, we get that this integral is $1$ so the rest follows from a middle value theorem.

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    $\begingroup$ I am unsure what this has to do with the question at hand. Can you elaborate a bit? $\endgroup$
    – Severin Schraven
    Commented Mar 8 at 17:04
  • $\begingroup$ @SeverinSchraven Presumable "middle value theorem" refers to the mean value theorem. If $D = [0,1]\times[0,1]$ and $\int_D g(x,y) dxdy$, then for some $(a,b) \in D$ we have $g(a,b) = \int_D g(x,y) dxdy/\int_D dxdy = 1$. Thus, $f(a)f(b)=1$. We may assume, WLOG, that $a \leq b$. However, I don't think it's guaranteed that $a < b$. $\endgroup$
    – Étienne Bézout
    Commented Mar 8 at 17:13
  • $\begingroup$ @ÉtienneBézout Thanks for the explanation. Indeed, I don't think we get $a<b$. $\endgroup$
    – Severin Schraven
    Commented Mar 8 at 17:20

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