Let $x\in \mathbb{R}^n$ and $f(x)\in \mathbb{R}^n$ be a smooth function. Consider a symmetric matrix $M(x)\in \mathbb{R}^{n\times n}$, whose elements are smooth. I am trying to prove the following (dont know if this holds):
$$f(x)^\top M(x) f(x)\leq 0 \iff M(x) \leq 0.$$
What have I tried:
Method(1)
We need to prove:
- Case(a): $ M(x) \leq 0 \implies f(x)^\top M(x) f(x)\leq 0$.
- Case(b): $f(x)^\top M(x) f(x)\leq 0 \implies M(x) \leq 0$.
Case (a) is trivial. In Case (b), if we consider the senariao when inequality holds strictly, then $f(x)^\top M(x) f(x)< 0 \implies M(x) < 0$. However, in the event the equality holds i.e., $f(x)^\top M(x) f(x)= 0$, then one or more of the following hold:
- $M(x)=0$
- $M(x)f(x) = 0$, i.e., $f(x)$ is in the left null space of $M(x)$.
- $M(x)^\top f(x) = 0$, i.e., $f(x)$ is in the right null space of $M(x)$.
I am stuck here.
- Method (2): Since $M(x)$ is symmetric, $LU$ decomposition holds with $L=U^\top$. Hence, let $M(x)=L^\top(x)D(x) L(x)$ with $D$ as a block diagonal matrix, then we have $$(L(x)f(x))^\top D(x) (L(x)f(x))\leq 0$$ Again I am stuck!
I constantly tell my students matrix language is an abomination when theory is what is to be understood, and you are the quintessential example of what I mean. Having said that, the exercise is either trivially false or trivially true.
False. If you mean that for some $f$ the inequality $f(x)^\intercal M(x) f^(x) \leq 0$ implies $M(x)$ to be negative semidefinite, consider the constant functions (hence smooth, even analytic) $M =\left[\begin{matrix}-1&0\\0&1\end{matrix}\right]$ and $f=(1, 0)^\intercal.$
True. If you mean that the relation $f(x)^\intercal M(x) f(x) \leq 0$ for all smooth functions $f$ implies $M(x)$ to be negative semidefinite, then simply consider $f$ constant functions and you obtain the definition of negative semidefinite.