Let $g\in L^{1}(\mathbb{R})$ Suppose that for every $h\in \mathbb{R}$
$$\int_{\mathbb{R}}|g(x+h)-g(x)|dx \leq |h| $$ Prove that $|yF(g)(y)| \leq 1 $ for all $y\in \mathbb{R} $ where F is the fourier transform.
Im quite confused by the above problem,
It looks like we have something close to the derivate if we bring h inside the integral, so maybe it is something like the mean value theorem? or is there something sneaky about the fourier transform i can use?
On
\begin{align} \hat{g}(y)(1-e^{ihy})&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(x)(e^{-ixy}-e^{-i(x-h)y})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(g(x)-g(x+h))e^{-ixy}dx \\ \left|y\hat{g}(y)\frac{1-e^{ihy}}{ihy}\right| & \le \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|g(x)-g(x+h)|dx \frac{1}{|h|} \\ & \le \frac{1}{\sqrt{2\pi}}. \end{align} Letting $h \downarrow 0$ gives $$ |y\hat{g}(y)| \le \frac{1}{\sqrt{2\pi}}. $$ If the author assumes the Fourier transform does not have the $\frac{1}{\sqrt{2\pi}}$ multiplier, then $|y\hat{g}(y)| \le 1$.
This is not a complete answer, but one can obtain an upper bound of $\frac{\pi}{2}$($\approx 1.5708$) by elementary methods as follows. Maybe this method could be useful to obtain the sharper bound that you are looking for.
First, write $$ \begin{align} \mathcal F(g)(y) &= \int_{\mathbb R} e^{-ixy}g(x)\mathrm d x \\ &= \int_{\mathbb R} e^{-i(x+h)y} g(x+h) \mathrm d x \\ &= e^{-ihy}\int_{\mathbb R} e^{-ixy} g(x+h) \mathrm dx \end{align} $$ If we choose $h$ such that $e^{-ihy}=-1$, that is $h=\frac{\pi}{y}$, we obtain $$ \begin{align} 2\mathcal F(g) (y) & = \int_{\mathbb R} e^{-ixy} \left (g(x)-g \left(x+\frac{\pi}{y} \right) \right)\mathrm d x \\ \implies 2|\mathcal F(g)(y)| &\le \int_{\mathbb R}\left |g(x)-g\left (x+\frac{\pi}{y} \right) \right | \mathrm d x \le \frac{\pi}{|y|} \end{align} $$ from which $$ |y\mathcal F (g)(y)|\le \frac{\pi}{2} $$ I do not know if the bound can be improved by following this path.