How to find $\max_{f}\int_{0}^{\infty}f\left(x\right)dx$ subject to $\int_{0}^{\infty}xf\left(x\right)dx=x_{0}$, where $f$ is a function and $x_{0}$ is a constant?
How to solve $\max_{f}\int_{0}^{\infty}f\left(x\right)dx$ subject to $\int_{0}^{\infty}xf\left(x\right)dx=x_{0}$?
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0@amu: If $f$ is a probability density function, then by definition $\int_0^{\infty} f(x) dx \leq 1$, where equality should be easily achievable by some appropriate choice of $f$. – 2012-02-29
2 Answers
Case 1: No restrictions on $f$
Taking $f(x)$ as a huge Dirac delta function centered at some small $\epsilon > 0$ as $f(x) = \frac{x_0}{\epsilon}\delta(x - \epsilon)$ gives you $\int_{0}^{\infty} x f(x) dx = x_0$ as required, and $\int_{0}^{\infty} f(x) dx = \frac{x_0}{\epsilon}.$ Letting $\epsilon \to 0$ thus gives you that the maximum is unbounded (unless you are given more restrictions on $f$).
Case 2: $f$ is a probability density function
Then, by definition, $\int_{0}^{\infty} f(x) dx \leq 1,$ and taking $f(x) = \delta(x - x_0)$ achieves this bound. Note that this $f$ corresponds to a discrete random variable $X$ with $P(X = x_0) = 1$.
For extra insight, note that the question can be nicely formulated in terms of probabilities:
Find $\max_F (1 - F(0))$ subject to $E(X) = x_0$.
Then the solution $X \equiv x_0$ described above becomes even more obvious.
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0@TMM, Really nice trick with the Dirac Delta. Any chance for assistance here: http://math.stackexchange.com/questions/1066495/minimization-of-variational-total-variation-tv-deblurring – 2014-12-14
Since $f$ is a probability density function, $\max_f \int_0^\infty f(x)\mathrm dx \leq \max_f \int_{-\infty}^\infty f(x)\mathrm dx = 1.$ Since $x_0$ is necessarily a positive number, then for any positive random variable $\int_0^\infty f(x)\mathrm dx = \int_{-\infty}^\infty f(x)\mathrm dx = 1$ (since $f(x) = 0$ for $x < 0$) and so $\max_f \int_0^\infty f(x)\mathrm dx = 1$.
Or, did you mean to ask
What is $\max_f \max_{x\in (0,\infty)} f(x)$ over all probability density functions $f(x)$ such that $\int_0^\infty xf(x)\mathrm dx = x_0$ where $x_0$ is a finite positive constant?
in which case TMM's answer shows that $f$ can have arbitrarily large value.
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0I see. Thank you so much! – 2012-03-01