Does the supremum $\sup_{t\in\mathbb R, |y|\le 1} |F(t, y)|$ happen to be finite?

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Assume $F:\mathbb R\times \mathbb R\to \mathbb R$ be a function of class $C^1$ which is $T$-periodic in $t$ (here $T>0$). Let $u\in L^2(\mathbb R)$.

During the last practice of my calc class, the lecturer stated the following: if we assume $\|u\|_{\infty} \leq 1$, then $$\int_{-T}^T F(t, u(t)) \, |u(t)|^2 dt \le \max_{t\in [0, T], |y|\le 1} |F(t, y)| \int_{-T}^T |u(t)|^2 dt,$$ with the maximum being finite. (Let us ignore the term $\int_{-T}^T |u(t)|^2$).

Now, assume that $F:\mathbb R\times \mathbb R^n\to \mathbb R$ (not $T$-periodic) and again $\|u\|_\infty \le 1 $ ad assume you want to do the same as before but for $$\int_{\mathbb R} F(t, u(t)) |u(t)|^2 dt.$$ How to do that? Which assumptions on $W$ are required?

It seems to me that (if is correct what I am writing) that $$\int_{\mathbb R} F(t, u(t)) |u(t)|^2 dt\le\sup_{t\in\mathbb R, |y|\le 1} |F(t, y)| \int_{\mathbb R} |u(t)|^2 dt$$ it is not necessarily finite. Which extra assumptions on $W$ are required?

I asked the lecturer and he replied "it is the same as before", but I am not convinced at all.

Anyone could please help?