Boundedness of Fourier transform Here we use Fourier transform $\hat{f} (\mu)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(x)e^{-ix\mu}dx$. we can prove that $\|\hat{f}\|_{L^{\infty}}\leq \frac{1}{2\pi}\|f\|_{L^{1}}$. Now the question is that can we find a Schwartz function $f$ such that the above inequality is strict and what's the sufficient and necessary condition for the equality holding.
Now for the first question, I tried Guassian function and characteristic function. And I all got equality. I don't know how to approach this question,  since we don't know too many functions for which we know their exact Fourier transform.
 A: HINT
This follows directly from the bound $$|\hat f(\mu)| \le \frac{1}{2\pi}\int_{-\infty}^\infty |f(x)e^{-ix\mu}|dx = \frac{1}{2\pi}\int_{-\infty}^\infty |f(x)|dx. $$
It is an equality for the Gaussian (or any positive function) since $$ \sup_\mu |\hat f(\mu)| \ge f(0) = \frac{1}{2\pi}\int_{-\infty}^\infty f(x)dx.$$
EDIT
Sorry, misread the question and re-marked this (non) answer as a hint since it does tell you that in order to find a strict inequality we must try a function that is not strictly positive. 
EDIT 2
Here's an example: $$f(x) = \frac{1}{2}x e^{-|x|}$$
We have $$ 2\pi \hat f(\mu) = \frac{i}{2}\int_{-\infty}^\infty xe^{-|x|}\sin(\mu x)\; dx = i\frac{2\mu}{(\mu^2+1)^2}$$ so $$2\pi|| \hat f(\mu)||_\infty = \frac{3\sqrt{3}}{8} < 1$$
whereas $$||f||_1 = \frac{1}{2}\int_{-\infty}^\infty |x|e^{-|x|} = 1.$$
A: 
Equality happens iff $f(x) = A e^{i \omega_0 x} |f(x)|$ (equality in $ L^1$) for some $\omega_0\in \mathbb{R}$ and $|A|= 1$.

If $f \in L^1$ then $\hat{f}(\omega)$ is continuous and $\to 0$ as $\omega \to \infty$. So if the equality happens it is at some finite $\omega_0$. Finally $$|\int_{-\infty}^\infty f(x) e^{-i \omega_0 x}dx| = \int_{-\infty}^\infty |f(x)| dx \implies f(x) = e^{i \omega_0 x} A |f(x)|$$ where $|A| = 1,A = \frac{f(x_0)e^{-i \omega_0 x_0}}{|f(x_0)|}$
