If $\int_{0}^{1} f dx=3$ and $\int_{0}^{1} xf dx =2$, then find the maximum value of $$\int_{0}^{1} f^2 dx.$$
What methods would apply to find this maximum value? I am not approaching the methods...
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Parvesh Kumar
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You can use the technique of calculus of variation. – Samrat Mukhopadhyay May 14 '15 at 05:13
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Sir if possible please explane.. – Parvesh Kumar May 14 '15 at 05:13
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@ParveshKumar Do you have any contraint on $f$? Continuity? $f\geq0$? – Daniel May 14 '15 at 05:22
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no othere conditions are given.. – Parvesh Kumar May 14 '15 at 05:23
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Without continuity, this is false. – matt biesecker May 14 '15 at 05:34
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There is no maximum. – copper.hat May 14 '15 at 05:36
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1One can show $\int_0^1 f^2 dx \ge 12$, though a maximum looks unlikely. – Macavity May 14 '15 at 05:43
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ok thanks a lot... its more than sufficient help.. – Parvesh Kumar May 14 '15 at 07:08
2 Answers
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Let $f(x) = 6x + \lambda (x^2-x+{1 \over 6})$. It is easy to check that $f$ satisfies the conditions and $\int_0^1 f^2 = 12+\lambda^2 {1 \over 180}$.
Hence we can choose $\lambda$ as large as we want and so there is no maximum.
Here is how I approached the problem: Let $\langle f, g \rangle = \int_0^1 f(x)g(x) dx$.
Then we are looking to maximise $\|f\|^2$ for $f$ such that $\langle f, x \mapsto 1 \rangle = 3$, $\langle f, x \mapsto x \rangle = 2$. By choosing a function that is orthogonal to $x \mapsto 1, x \mapsto x$, we can add multiples of this function without changing the value of the constraints. In this case, one such function is $x \mapsto x^2-x+{1 \over 6}$. The rest is algebra.
copper.hat
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really nice explanation - great connection with the interpretation of the constraints as orthogonal basis elements. – izœc May 14 '15 at 10:47
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For a comprehensive introduction to the method of calculus of variation, you may look at this reference.
We want to maximize $\int_{0}^1 f^2 dx$ subject to the conditions $\int_{0}^1 f dx=3,\ \int_{0}^1 xf dx=2$. We form the Lagrangian, $l(x,\lambda,\mu)=f^2(x)-\lambda f(x)-\mu xf(x)$. Then use the Euler-Lagrange equations to get $$\frac{\partial l}{\partial x}=0\implies 2f(x)f'(x)-\lambda f'(x)-\mu(f(x)+xf'(x))=0$$ Solve this differential equation for $f(x)$ and use the constraints to solve for $\lambda,\mu$ and that will give you an extremum.
Samrat Mukhopadhyay
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