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When does the minimum of: $\displaystyle{\int_0^1 f^2(x) dx - 2\lambda \int_0^1 x f(x) dx + \frac{\lambda^2}{3}}$ occur?

I have no clue, other than this looks like a quadratic.

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1 Answer 1

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It looks like a quadratic because it is a quadratic: \begin{align} 0\le\int_0^1(f-\lambda x)^2&=\int_0^1f^2-2\lambda\int_0^1xf+\lambda^2\int_0^1x^2\\ &=\int_0^1f^2-2\lambda\int_0^1xf+\lambda^2\cdot \frac{1}{3}. \end{align}

So the minimum is $0$ and it is attained when $f(x)=\lambda x$.

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  • $\begingroup$ It is actually possible to explicitly say what $\lambda$ here should be in the integral form. $\endgroup$
    – Naz
    Dec 2, 2015 at 15:10

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