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Is there any continuous function $y=f(x)$ on $[0,1]$ that satisfies the following conditions? $$ f(0) = f(1) = 0, $$ and $$ f(a)^2-f(2a)f(b)+f(b)^2<0, $$ for some $0

I tried to test with several functions (with different $a,b$) but non of them satisfied.

Any help is appreciated. Thank you in advance.

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    is this for *any* $0 – 2017-01-17
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    What is the domain of $f$? – 2017-01-17
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    For any $0 – 2017-01-17

2 Answers 2

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What you want is a function that is highly peaked around $x=2a$. And you need to enforce the conditions that $f(0)=f(1)=0$; for that you can put in a factor of $x(1-x)$.

So try

$$ f(x) = x(1-x)e^{-20(x-\frac12)^2} $$ with $a=\frac14$ and $b=\frac34$.

There, $$f(a)^2 - f(2a)f(b) + f(b)^2 \approx -0.00766 <0$$

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    Thank you very much for your help! – 2017-01-17
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All you have to do is choose numbers $a,b$ so that $0\lt a\lt 2a\lt b\lt1$ (or $0\lt a\lt b\lt2a\lt1$), then find numbers $y_1,y_2,y_3$ satisfying $y_1^2-y_2y_3+y_3^2\lt0,$ and then construct a continuous function (e.g. a fourth degree polynomial) $f(x)$ with $f(0)=f(1)=0,\ f(a)=y_1,\ f(2a)=y_2,\ f(b)=y_3.$ For example:

$$f(x)=x(x-1)(4x-1)(8x-7),\ a=\frac14,\ b=\frac34$$

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    Nice example! Thank you very much! – 2017-01-18
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    @AnaUspekova: Or you can just draw straight lines to connect the dots required. – 2017-01-18
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    Is there a general approach to solving a problem of this kind? i.e. Find a function f that obeys the following constraints. – 2017-01-18
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    ....or even more specifically: Find a polynomial function that satisfies the following constraints. – 2017-01-18
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    @curious_cat Are you asking about the [Lagrange interpolation formula](http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html) or something else? – 2017-01-18
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    @bof Thanks. No, I meant is there a more algorithmic way to come up with such functions given constraints. – 2017-01-18
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    @curious_cat I have no idea what kind of functions you want to come up with or what kind of constraints you want them to satisfy. Maybe if you could give an example or two I would at least understand the question, though I don't expect to know the answer. – 2017-01-18