Let $f\in C^\infty(\mathbb{R})$ be a smooth function, and assume its (distributional) Fourier transform $\hat{f}$ is smooth away from 0, and that $\hat{f}$ is rapidly decaying at infinity. Choose some $\varepsilon >0$, and a function $\varphi \in C^\infty (\mathbb{R})$, $\varphi \geq 0$, such that $\varphi \equiv 0$ on $(-\frac{\varepsilon}{2},\frac{\varepsilon}{2})$ and $\varphi \equiv 1$ outside $(-\varepsilon ,\varepsilon)$. Define $f_\varepsilon$ via $\hat{f_\varepsilon}=\varphi \hat{f}$. This is a Schwartz function.
Question: Is there an upper bound for $||f_\varepsilon||_\infty$ which is independent of $\varepsilon$?
Feel free to assume $f$ to be from the function class $$ S^0_1(\mathbb{R}):= \left\lbrace f\in C^\infty(\mathbb{R}) : |f^{(j)}(x)|\leq C_j (1+|x|)^{-j} \right\rbrace. $$ I would even be perfectly happy with an answer for the specific function $f(x)=x(x^2+1)^{-1/2}$.
My thoughts so far: The supremum norm of $ f_\varepsilon $ may be estimated (up to some multiplicative $ \pi $-factor I will surpress) via \begin{align} || f _\varepsilon || _\infty = \sup_x \left| \int e^{i\xi x} \hat{f} _\varepsilon(\xi) d\xi \right| \leq \int |\hat{f} _\varepsilon (\xi)| d\xi =||f _\varepsilon||_1 . \end{align}
Intuitively, $\hat{ f_\varepsilon } $ should converge to $ \hat{f} $ pointwise away from 0. Because $ \hat{f} $ may explode at 0, I see no reason to expect $|| \hat{f} _\varepsilon ||_1$ to stay bounded for $ \varepsilon \to 0 $. On the other hand, the above estimate is quite rough, and $|| \hat{f} _\varepsilon ||_1$ might not reflect the behavior of $ ||f _\varepsilon || _\infty $ at all.
I would appreciate any insights into this problem!
I am afraid that the inequality you got, namely $$ \lVert f_\epsilon\rVert_\infty\le \lVert \widehat{f}_\epsilon\rVert_1$$ is all you can get. This inequality is sharp in the sense that there are functions $f$ such that it holds with equality sign.
Indeed, if $f$ is such that $\widehat{f}_\epsilon$ is:
then $$ \frac12 \lVert f_\epsilon \rVert_\infty = \sup_{x\ge 0} \left\lvert \int_0^\infty \widehat{f}_\epsilon(\xi)\cos(x\xi)\, d\xi\right\rvert = \int_0^\infty \widehat{f}_\epsilon(\xi)\, d\xi =\frac12 \lVert \widehat{f}_\epsilon\rVert_1.$$
The key point here is of course the second equality. You can prove it by observing $$\tag{1} \int_0^\infty \widehat{f}_\epsilon(\xi)(1\pm\cos(x\xi))\, d\xi \ge 0$$ which readily implies what we want, namely, that $\sup_{x\ge 0} \left\lvert \int_0^\infty \widehat{f}_\epsilon(\xi)\cos(x\xi)\, d\xi\right\rvert = \int_0^\infty \widehat{f}_\epsilon(\xi)\, d\xi$.
EDIT. In the previous version of this answer I required a third condition: that $\widehat{f}_\epsilon(\xi)$ were decreasing for $\xi \ge 0$. That condition is not needed and I removed it.