Let $\alpha: \mathbb{R}_{\geq 0} \times \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ be a $\mathcal{KL}$-function. For each fixed $t$, the continuous function $\alpha(x,t)$ belongs to class $\mathcal{K}$, i.e., a strictly increasing function such that $\alpha(0,t) = 0$ and for each fixed $x$, the function $\alpha(x,t)$ is decreasing with respect to $t$ and $\alpha(x,t) \to 0$, as $t \to \infty$. Also, let $b \in \mathbb{R}_{\geq 0}$. Is there any way to get a lower bound using a class $\mathcal{K}$ function $\alpha_1(x)$ or another class $\mathcal{KL}$ function $\alpha_2(x,t)$ in the sense of the following:
$$ \alpha(x,t) + b \geq \alpha_1(x + b), \quad \forall t \geq 0. $$
or
$$ \alpha(x,t) + b \geq \alpha_2(x + b,t), \quad \forall t \geq 0. $$
I understand that there is a similar question for class $\mathcal{K}$ functions, but I was wondering if the argument would hold similarly for class $\mathcal{KL}$ functions and hence, I am asking this as a separate question.
Let $\alpha$ be a class-$\mathcal KL$ function and $b\ge0$.
Lower bound with a class-$\mathcal K$ function
We're looking for a class-$\mathcal K$ function $\alpha_1$ such that $$\forall x,t\ge 0,\,\alpha(x,t)+b\ge\alpha_1(x+b).$$
Taking the limit $t\to +\infty$, we see that $\alpha_1$ must be bounded by $b$. On the other hand, $x\to\alpha(x,t)$ being increasing for each $t\ge0$, we have $\alpha+b\ge b$. Thus any class-$\mathcal K$ function $\alpha_1$ bounded by $b$ will do the work (for example $x\mapsto\frac{2b}{\pi}\arctan(x)$).
Lower bound with a class-$\mathcal KL$ function
We're looking for a class-$\mathcal KL$ function $\alpha_1$ such that $$\forall x,t\ge 0,\,\alpha(x,t)+b\ge\alpha_1(x+b,t).$$
We will try to modify the construction in the answer for the question in the class-$\mathcal K$ case to ensure that $\alpha_1$ belongs to $\mathcal KL$:
You see that merely taking $\alpha_1(x,t)=\alpha(x-b,t)+b$ doesn't solve the problem: we lack the condition $\alpha_1\to 0$ when $t\to +\infty$. Multiplying by something that goes to $0$, like $e^{-t}$ does the work. Furthurmore, $e^{-t}\le 1$ for $t\ge 0$, so it doesn't negatively affect our desired inequality. Hence we define $$\forall x\ge b,\forall t\ge 0,\alpha_1(x,t)=(\alpha(x-b,t)+b)e^{-t}.$$
Finally, we should define $\alpha_1$ on $[0,b]\times\mathbb R_{\ge 0}$. Notice that $\alpha_1(b,t)=be^{-t}$, so any function continuous increasing function $x\mapsto f(x)$ such that $f(0)=0$ and $f(b)=b$ will do. You can take $$\alpha_1(x,t)=xe^{-t},\,\,\,\,x\in[0,b],t\ge 0.$$