Do you think integration by parts will work?
Continuous function f : [ 0 , 1 ] → R satisfies following conditions
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ∫ 0 1 f ( x ) d x = 1 ∫ 0 1 x f ( x ) d x = 1 ∫ 0 1 ( f ( x ) ) 2 d x = 4
Evaluate ∫ 0 1 ( f ( x ) ) 3 d x .
The answer is 10.
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1 solution
Nice illustration in proving that f ( x ) = 6 x − 2 for 0 ⩽ x ⩽ 1 , but is this a unique function of f ( x ) ? Can we prove it?
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Since I have shown that ∫ 0 1 ( f ( x ) − 6 x + 2 ) 2 d x = 0 that proves (since f is continuous) that f ( x ) = 6 x − 2 .
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Oh I missed that constraint (continuous part), I was thinking that I could come up with some "expression with floor function".
A blunder on my part. Thanks for your swift response as usual.
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@Pi Han Goh – If we dropped the continuity condition, the worst that could happen is that f ( x ) = 6 x − 2 almost everywhere (the identity might fail on a set of Lebesgue measure zero). This would not affect the value of the integral of f 3 .
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@Mark Hennings – I understood this as well. Wow, it took me less than 3 years to finally understand this statement.
You have my utmost respect.
I think you skipped a step. To prove it's the only solution you have to show that the integrand is non-negative on the interval, so must be zero everywhere. That wasn't immediately obvious.
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@Ilya Lyubarsky – No I didn't. The square of a real-valued function is non-negative, which is obvious.
Note that ∫ 0 1 ( f ( x ) − A − B x ) 2 d x = A 2 + 3 1 B 2 + A B − 2 A − 2 B + 4 = ( A + 2 ) 2 + 3 1 ( B − 6 ) 2 + ( A + 2 ) ( B − 6 ) Choosing A = − 2 , B = 6 tells us that ∫ 0 1 ( f ( x ) + 2 − 6 x ) 2 d x = 0 so that f ( x ) = 6 x − 2 for 0 ≤ x ≤ 1 . Thus ∫ 0 1 f ( x ) 3 d x = ∫ 0 1 ( 6 x − 2 ) 3 d x = 1 0