Does this infimum converge to $+\infty$?

Question

Consider a sequence of $\{ \phi^n\} \subset \mathcal{C}^2(\mathbb{R})$ such that \begin{align} &\left|\frac{\partial}{\partial x} \phi^{n}(x)\right| \leq M \\ &\lim _{n \rightarrow \infty} \frac{\partial^{2}}{\partial x^{2}} \phi^{n}(x)=+\infty \end{align} for some $M>0$.
Let $U\subset \mathbb{R}$ and consider $b, \sigma, l \colon \mathbb{R} \times U \to \mathbb{R} $ and consider for FIXED $x \in \mathbb{R}$ $$\inf _{u \in U}\left\{ b(x, u)\frac{\partial}{\partial x} \phi^{n}(x)+\frac{1}{2}\left|\sigma\left(x, u\right)\right|^{2} \frac{\partial^{2} \phi^{n}(x)}{\partial x^{2}}+l\left(x, u\right)\right\}$$ Now if you assume $|b(x,u)| \leq C$ for some $C>0$, $|\sigma(x,u)|>L$ for some $L>0$ and $l(x,u)$ bounded from below i.e. $l(x,u)>K$ for some $K \in \mathbb{R}$ then we have thanks to the properties of $\phi^n$ \begin{align}& \inf _{u \in U}\left\{ b(x, u)\frac{\partial}{\partial x} \phi^{n}(x) +\frac{1}{2}\left|\sigma\left(x, u\right)\right|^{2} \frac{\partial^{2} \phi^{n}(x)}{\partial x^{2}}+l\left(x, u\right)\right\}\\ & \geq \frac{\partial}{\partial x} \phi^{n}(x) \inf _{u \in U}\left\{ b(x, u)\right \} +\frac{1}{2} \frac{\partial^{2} \phi^{n}(x)}{\partial x^{2}} \inf _{u \in U}\left\{ \left|\sigma\left(x, u\right)\right|^{2} \right \} + \inf _{u \in U}\left\{l\left(x, u\right)\right\}\\ & \geq -M C +\frac{1}{2}L \frac{\partial^{2} \phi^{n}(x)}{\partial x^{2}} + K \to +\infty \end{align} as $n \to \infty$, right?
Are there weaker assumptions under which the same is true?
Note that if $b,\sigma,l$ are continuous and $U$ is compact then the infimums are minimums and then the conditions $|b(x,u)| \leq C$ and $l(x,u)>K$ can be removed and we need to assume only that $\sigma(x,u) \neq 0$.

Answer 1

Since you fix $x$, the problem can be reduced to :

$$ \inf_{u \in U} \{ b_x(u) x_n + \frac{1}{2} |\sigma_x(u)|^2 y_n + l_x(u) \} $$

where $b_x, \sigma_x, l_x$ are the functions $b, \sigma, l$ restricted to $\{ x \} \times \mathbb{R}$, and (in broad generality) $x_n$ is a bounded sequence, $y_n$ is a sequence converging to $+\infty$. I drop the $x$ in the following.

You can choose $x_n$, $y_n$ arbitrarily at a given point $x$, for instance through :

$$ \phi^n(y) = \frac{y_n}{2} (y - x)^2 + x_n (y - x) $$

The condition $\sigma(u) \neq 0$ for every $u$ is clearly necessary and I assume it from now on. I assume that the assumptions you suggested are taken uniformly for $u$ at a fixed point $x$ (i.e. for instance $|b(x, u)| \leq C$ means that, for the fixed $x$ considered, $\sup_u |b(x, u)| \leq C$).

You can actually choose some slightly lighter assumptions by coupling them. For instance, it is not needed for $l$ to be uniformly bounded by below as long as you have a control of the form :

$$ C |\sigma(u)|^2 \geq (l(u))_{-} $$

for some constant $C > 0$, uniformly in $u$, where $a_{-}$ denotes the negative part of $a$. Indeed, in this case, you have as $n \to +\infty$ that $(l(u))_{-} = o(|\sigma(u)|^2 y_n)$ uniformly in $u$.

It can be shown that this first condition on $l$ is necessary (by contradiction, by choosing $u_n$ such that $n |\sigma(u_n)|^2 < -l(u_n)$, $x_n = -\varepsilon(b(u_n))$ where $\varepsilon$ is the sign function, $y_n = \ln(n)$ for instance and using the fact that the infimum is less than the value at $u_n$).

The control of $b$ is a little bit different, since the behaviour of $b$ can also be compensated by some nice behaviour of $l$, namely : if $\sigma$ gets close to $0$ it can be corrected with $l$ going to $+\infty$. You may therefore ask for the following : there exists a function $\omega : \mathbb{R}^{+*} \to \mathbb{R}$ such that :

1. $\lim_{x \to +\infty} \omega(x) = +\infty$
2. $\omega$ is bounded by below
3. for every $u$, $(l(u))_{+} - M |b(u)| \geq \omega(|\sigma(u)|^{-1})$.

This condition, the non-nullity condition on $\sigma$ and the condition given above for the control of the negative part of $l$ are sufficient. Indeed, you have that :

$$ b(u) x_n + \frac{1}{2} |\sigma(u)|^2 y_n + l(u) \geq (l(u))_{+} - M|b(u)| + \frac{1}{2} |\sigma(u)|^2 y_n + (l(u))_{-} \\ \geq \omega(|\sigma(u)|^{-1}) + \frac{1}{2} |\sigma(u)|^2 y_n - C |\sigma(u)|^2 $$

Assume that the limit in $n$ of the infimum in $u$ of the right-hand side is not $+\infty$. Then you may find $u_n$ such that $\omega(|\sigma(u_n)|^{-1}) + \frac{1}{2} |\sigma(u_n)|^2 y_n - C |\sigma(u_n)|^2$ is bounded by above. Since $\omega$ is bounded by below and $y_n \to +\infty$, this implies that $\sigma(u_n) \to 0$, but this would imply $\omega(|\sigma(u_n)|^{-1}) \to +\infty$, a contradiction.

I believe that the existence of such an $\omega$ is necessary - but I didn't try to prove it so far (it is probably quite difficult). Note that, under your assumptions, the existence of such an $\omega$ is straight-forward. However, it already provides a set of conditions far less restricting than your initial ones (but less easy to manipulate !) The main intent of my answer is to show that, by controlling the behaviour of $\sigma, b, l$ together, you may control the infimum with less restrictive conditions.