# Non-negative integral implies non-negative integrand (up to total derivatives)?

Take a functional $S[\varphi]$ on a $d$-dimensional space-time of the form

$S[\varphi]=\int d^d x\, L(x,\varphi,\partial_\mu \varphi,\partial_\mu \partial_\nu \varphi, \dots)$

where $\varphi(x)$ is a scalar function of $x^\mu$, and $L$ is a polynomial function of $\varphi$ and its derivatives, with $x^\mu$-dependent tensor-valued coefficients, and contains only even powers in $\varphi$ e.g.

$L = c_1(x) \varphi^2+ c^\mu_2(x)\varphi^3 \partial_\mu \varphi+c_3(x)^{\mu\nu}\partial_\mu \varphi \partial_\nu \varphi+c_4(x)^{\mu\nu}\varphi^3 \partial_\mu\partial_\nu \varphi+\cdots\ .$

Can one show that

$S[\varphi] \geq 0 \quad\text{for all }\varphi\qquad \Rightarrow\qquad L+\partial_\mu K^\mu\geq 0\quad\text{for all }\varphi ,$

where we consider configurations of $\varphi$ such that $\varphi$ and its derivatives vanish sufficiently fast at space and time infinity, and where $K^\mu$ is some vector that depends on $x^\mu$, $\varphi$ and the derivatives of $\varphi$?

In case of negative answer, is there any hope for the statement to be true if one restricts to more specific forms of $S[\varphi]$ (e.g. if one assumes the coefficients $c_1,\ c_2^\mu,\ \dots$ to be independent of $x^\mu$)?

## migrated from physics.stackexchange.comFeb 2 '17 at 4:48

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• I'm not entirely sure what you want to do here. The statement obviously holds when $L$ has only even powers of $\varphi$, and I'm having trouble imagining an $L$ such that $S[\varphi]\geq 0$ holds where not already $L\geq 0$ without adding anything. Do you have an example where $L\geq 0$ does not already hold but $S \geq 0$ does? – ACuriousMind Jan 31 '17 at 14:02
• I'm sorry, I realized that some of the conditions I put in the original post were not right. I edited the question in two points: first, $L$ can have tensor-valued coefficients (see the explicit example of $L$ above), and second, the configurations of $\varphi$ that we consider are such that $\varphi$ and its derivatives vanish sufficiently fast at space and time infinity. Regarding the example, I can then of course consider $L=\partial_\mu K^\mu$, where $K^\mu=c^\mu \varphi^2$, so that $S=0$ but clearly $L$ does not have a definite sign. – vvega Jan 31 '17 at 17:25
• Not that it's not interesting, but this is really a math's question, isn't it? – pppqqq Jan 31 '17 at 18:05
• I see your point pppqqq, I thought about posting it on math too, but given the many posts on Lagrangian formalism and total derivatives in physics I decided to ask the question here. – vvega Jan 31 '17 at 18:23
• Comment to the question (v4): The assumption that the action is non-negative $\forall V \subseteq \mathbb{R}^n\forall \phi :~~ S_V[\phi]\equiv \int_V \! d^nx ~{\cal L}~\geq~ 0$ implies that the Lagrangian density ${\cal L}\geq 0$ is non-negative by localization of $V$. – Qmechanic Jan 31 '17 at 21:48