Answer :

Solution :

Given function is even,

${\displaystyle \frac{dF(\omega )}{d\omega }}=-\omega F(\omega )$.... (1)

From differentiation property,

$tf\left(t\right)=\frac{jdF\left(\omega \right)}{d\omega }$

Applying IFT to the above equation,

$-jtf\left(t\right)=\frac{jdf\left(t\right)}{df}$

$\frac{df\left(t\right)}{dt}=-tf\left(t\right)$.... (2)

From equation (1) and (2) it is clear that the f (t) is the change of Gaussian function, it can be written as,

$f\left(t\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-t^2}{2}}$.... (3)

From equation (3),

$\frac{1}{\sqrt{2\pi }}{e}^{\frac{-t^2}{2}}\leftrightarrow e^{\frac{-{\omega }^2}{2}}$

$f\left(t\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-t^2}{2}}$

$f\left(0\right)=\frac{1}{\sqrt{2\pi }}$

$f\left(0\right)=0.3989$

$\therefore \mathbf{f}\left(0\right)<1$