Does the pointwise identity $\frac{e^{tf}-1}t\xrightarrow{t\to0+}f$ also hold with respect to the supremum norm? Let $f\in C(\mathbb R^d;\mathbb C)$ and $g\in C_b(\mathbb R^d;\mathbb C)$. Are we able to show that $$\frac{e^{tf}g-g}t\xrightarrow{t\to0+}fg\tag1$$ with respect to the supremum norm on $C_b\mathbb R^d;\mathbb C)$?
Since $$\lim_{t\to0+}\frac{e^{tc}-1}t=\left.\frac{\rm d}{{\rm d}t}e^{tc}\right|_{t=0}=c\tag2,$$ $(1)$ obviously holds pointwisely.
And since the supremum norm is clearly submultiplicative, it should be sufficient to show $$\left\|\frac{e^{tf}-1}t-f\right\|_\infty\xrightarrow{t\to0+}0\tag3.$$
Are we able to show this?
Remark: In my application, I know that $|f(x)|\le c(1+|x|^2)$ for all $x\in\mathbb R^d$ for some $c\ge0$. So, if necessary, feel free to assume this.
Remark 2: What I'm trying to understand is the equation after (6.5) in this paper.
 A: If $f(x)$ is a function of polynomial growth, then $(e^{tf(x)}-1)/t$ converges to $f(x)$ uniformly on compact sets of $x$, but it does not need to converge uniformly as $t\to 0$. Multiplying by a function $g(x)$ could in principle help the convergence, but $g(x)$ would have to decrease to $0$ very fast. For example, if $|f(x)|=O(|x|^2)$ as $|x|\to\infty$, then $g(x)$ should decay even faster than $e^{-|x|^2}$, otherwise for any fixed $t$, we could make the expression
$$\frac{e^{tf(x)}g(x)-g(x)}{t}$$
have size comparable to $1$ for $x$ very large.
In the document you linked to, $g(x)$ is a Schwartz function, so we know that $g(x)$ does decay faster than any fixed polynomial, which seems useful, but because your $f(x)$ grows like a polynomial, $g(x)$ being Schwartz isn't enough to conclude that the convergence you are aiming for is uniform, because decaying faster than any fixed polynomial is not the same as decaying faster than $\exp(-|\text{polynomial}|)$.
Now I'll switch to the notation used in the document you linked to. Presumably, you are interested in the question about uniform convergence because you want to take the inverse Fourier transform of $\frac{\widehat{P_tf}-\widehat f}{t}$ and argue that as $t\to 0$, by uniform convergence, this would converge to $\mathscr F^{-1}(-\psi \widehat f)$. Equivalently, you want to exchange the derivative and the integral in
\begin{align*}
\int e^{ix\cdot\xi}\frac{\partial}{\partial t}\bigg|_{t=0}\widehat{P_tf}(\xi)\,d\xi &= \frac{d}{dt}\bigg|_{t=0}\int e^{ix\cdot\xi}\widehat{P_tf}(\xi)\,d\xi
\end{align*}
Fortunately, you don't need uniform convergence in order to justify this exchange. Using "the differentiation lemma for parameter-dependent integrals" that the author mentions is probably all you need, though I admit I don't have either of the given references on hand. Here is one formulation that works for what you need:

Differentating under the integral sign. Suppose that $F(x,t)$ is integrable as a function of $x \in \mathbb{R}^d$ for each value of $t \in \mathbb{R}$ and differentiable as a function of $t$ for each value of $x$.  Assume also that
$$\bigg| \frac{\partial}{\partial t} F(x,t) \bigg| \le G(x),$$
for all $x,t$, where $G(x)$ is an integrable function of $x$.  Then $\frac{\partial}{\partial t} F(x,t)$ is integrable as a function of $x$ for each $t$ and
$$\frac{d}{dt} \int F(x,t)\, dx = \int \frac{\partial}{\partial t} F(x,t)\,dx.$$

The proof is by the dominated convergence theorem, which is more flexible than the result that allows you to exchange limits and integrals under the condition of uniform convergence, as this problem you're having shows.
I will leave the proof of this formulation up to you, but in order to apply it, all we need to verify is that $-e^{ix\cdot\xi}\psi(\xi)\widehat{f}(\xi)$ is bounded by an integrable function of $\xi$. Since $f$ is Schwartz, $\widehat f$ is Schwartz too, so since $\psi$ only has fixed polynomial growth and $|e^{ix\cdot \xi}| = 1$, $|\!-\!e^{ix\cdot\xi}\psi(\xi)\widehat{f}(\xi)|\le C(1+|\xi|)^{-100d}$, where $d$ is the dimension of the ambient space and $C$ is some possibly large but fixed constant. The function $C(1+|\xi|)^{-100d}$ is integrable on $\mathbb R^d$, so we are done.
A: I don't understand your remark as $f$ being a bounded function already implies the bound $\vert f(x) \vert \leq c(1 + \vert x \vert^2)$ with $c = \Vert f \Vert_{\infty}$.
In spite of that, we have
$$
\frac{e^{t f(x)} - 1}{t} = f(x) \, e^{\theta f(x)}
$$ for a $\theta \in (0, t)$ by the Mean Value Theorem.
Therefore,
$$
\left\vert \frac{e^{t f(x)} - 1}{t} - f(x) \right\vert
= \vert f(x) \vert \, \vert e^{\theta f(x)} - 1 \vert
\, \leq \, \Vert f \Vert_{\infty} (e^{\vert\theta f(x) \vert} - 1)
\, \leq \, \Vert f \Vert_{\infty}(e^{t \Vert f \Vert_{\infty}} - 1),
$$ where the last term goes to zero as $t \to 0$, uniformly in $x$.
A: It seems to me that you can't have a uniform convergence, even adding the bound: let $f(x)=x$, which respect $|f(x)|\le (1+x^2)$, and $g=1$, which is bounded, then
\begin{align*}
\frac{e^{tx}-1}{t}\to+\infty && x\to+\infty,
\end{align*}
that contradicts the fact that the convergence is uniform.
