# Euler-Lagrange equation has no solution

I need to find a function $f(x)$ that maximizes a functional: $$J(f)= \int\limits_{-\infty}^{+\infty} e^{-x^2/2}f(x) \,dx$$

Where $$f(x)>0 \ \text{ and} \int\limits_{-\infty}^{+\infty} f(x) \,dx = 1$$ Euler-Langrange equation will simply be:

$$e^{-x^2/2}=0$$

So does it mean that there are no stationary points for this functional? But how do I search for a maxima/minima then?

Thank you!

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Indeed, the EL equations are a necessary condition for extremizing the functional, so it is immediate there are no $f$ that maximises your function. –  Ragib Zaman Sep 15 '12 at 10:31

The maximum cannot be attained. You're trying to maximize integrating a positive $f$ of mass 1 against $e^{-x^2/2}$, so this means you want to put as much of the mass of $f$ as possible on the large values of $e^{-x^2/2}$; of course this function is strictly maximized at $x = 0$. But then this means you can construct a sequence of functions $f_n(x)$ of increasing sharper peaks about $x = 0$ so that $J(f_n)$ is increasing towards 1; without obsessing over rigor, you can think of this sequence as weakly approaching the Dirac delta function, and one can observe that the upper bound for your functional is 1, which cannot be attained.
We can use probability: we have to find the maximum of $E[e^{-X^2/2}]$ where $X$ is a continuous random variable. We try to work with normal laws. Let $f_n(x):=\frac n{\sqrt{2\pi}}\exp\left(-\frac{x^2n^2}2\right)$, then $$\int_{-\infty}^{+\infty}f_n(x)\exp\left(-\frac{x^2}2\right)dx=\frac n{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\exp\left(-\frac{x^2}2(n^2+1)\right)dx.$$ Using the substitution $t=\sqrt{n^2+1}x$, we get $$\int_{-\infty}^{+\infty}f_n(x)\exp\left(-\frac{x^2}2\right)dx=\frac n{\sqrt{n^2+1}},$$ which proves that the maximum is $1$. It's never attained, otherwise we would have $$\int_{-\infty}^{+\infty}f(x)(1-e^{-x^2/2})dx=0$$ for some $f>0$, hence $f(x)(1-e^{-x^2/2})=0$ almost everywhere, which is not possible.
The $\delta$-distributon would match, if it is allowed in your search range. Otherwise, I assume you additionally require $f$ to be e.g. continuous. Let $f$ be a continuous function with $f(x)\ge0$ for all $x$ and $\int_{-\infty}^\infty f(x)dx=1$. Then $f(x)\ne0$ for some $x\in\mathbb R$ and by continuity $f(x_0)\ne0$ for some $x_0\ne0$. Let $h=f(x_0)$. Then there is $\epsilon>0$ such that $\epsilon<\frac12|x_0|$ and $f(x)>\frac h 2$ for all $x$ with $|x-x_0|<\epsilon$. Let $$g(x)=\begin{cases}-(x-x_0-\epsilon)(x-x_0+\epsilon)&\mathrm{if\ }x_0-\epsilon<x<x_0+\epsilon\\ (x-\epsilon)(x+\epsilon)&\mathrm{if\ }-\epsilon<x<+\epsilon\\ 0&\mathrm{otherwise}\end{cases}$$ and verify that $g$ is continuous, $\int_{-\infty}^\infty g(x) dx=0$, $J(g)>0$ and $f+c\cdot g\ge 0$ as long as $0\le c \le \frac h{2\epsilon^2}$.