Let $S$ be the set of all integrable on $[0,1]$ such that $$\int\limits_0^1f(x)dx=\int\limits_0^1xf(x)dx+1=3.$$ Prove that $S$ is infinite and evaluate $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$
Minimum of integral
4
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calculus
integration
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1To show that $S$ is infinite, just find two elements of $S$; then any linear combination is also a solution. For the minimum, the Euler-Lagrange equations with Lagrange multipliers say that $f$ should be linear, and you can determine the coefficients from the conditions; then you need to show that this is minimal in the absence of the differentiability assumptions of the Euler-Lagrange equations. – 2012-03-16
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0To show that the solution $f_0$ of the Euler-Lagrange equation is minimal even in absence of the differentiability assumption, write $f = f_0 + g$ and minimize $\int f(x)^2\,dx$ with respect to $g$ under the condition $\int g(x)\, dx= \int x g(x)\, dx =0$. – 2012-03-16