Let $f(x,y)\in \mathcal{C}([a,b]\times[c,d])$ such that $$\exists \xi\in (a,b) : f(\xi,y)\neq 0, \forall y\in [c,d].$$
By the continuity of $f$, we have $$|f(\xi,\cdot)|\geq \min\limits_{[c,d]} |f(\xi,\cdot)|=\alpha_0>0. \;(1)$$
I want to know if we can prove the following result:
There exists an open interval $I\subset [a,b]$ and $C>0$ such that $$\forall (x,y)\in I\times [c,d]: |f(x,y)|\geq C>0, \; (2).$$ If we use continuity of $f(\cdot,y)$ for each $y\in [c,d]$, by $(1)$ there exists an open interval $I_y$ containing $\xi$ such that $$\forall y\in [c,d], \forall x\in I_y: |f(x,y)|\geq \alpha_0>0.$$ But this is not what I'm looking for. I need two independent intervals such that $(2)$ holds.
If this is not true in general, can someone give a counter-example ?
You have $|f|\geq \alpha_0$ on $\{\xi\}\times [c,d].$ By the uniform continuity of $|f|$ on $[a,b]\times [c,d],$ there exists $\delta >0$ such that $|p-q|<\delta$ implies $| \,|f(p)|-|f(q)|\,|<\alpha_0/2.$ Thus $|\,|f(x,y)|-|f(\xi,y)|\,| < \alpha_0/2$ if $x\in (\xi-\delta, \xi+ \delta).$ It follows that $|f| > \alpha_0/2$ on $(\xi-\delta, \xi+ \delta)\times [c,d].$