For $f\in L^1(\mathbb{R}^n)$, we define $\hat{f}$ for $\alpha\in \mathbb{R}^n $ as $$\hat{f}(\alpha)=\int_{\mathbb{R^n}} f(x)e^{-i\langle x,\alpha\rangle}\,dx$$ Clearly $||\hat{f}||_{\infty}\le ||f||_1$. I want to know when does equality hold. Clearly if $f$ is non negative $\hat{f}(\textbf{0})=||f||_1$. So the equality holds. Do there exist a $f$ for which $\hat{f}(\textbf{0})\neq ||f||_1$ but still equality holds in above inequality?
I don't know if such a function really exists. Any hint/Suggestion will be appreciated.
$$f(x) = e^{-x^2} e^{ix}$$
$$\hat f(0) = \int_\mathbb R e^{-x^2} e^{ix} dx = \frac {\sqrt \pi }{\sqrt[4]e}$$
$$||f||_1 = \int_\mathbb R e^{-x^2} dx = \sqrt \pi$$
so $\hat f(0) \neq ||f||_1$
But since $\hat f(\alpha) = \int_\mathbb R e^{-x^2}e^{ix - i\alpha x}dx$, setting $\alpha = 1$ we get
$$\hat f(1) = \int_\mathbb R e^{-x^2} dx = \sqrt \pi$$
and from this we deduce $$||\hat f||_\infty = ||f||_1$$