Is $ H\left\{x_h(t)\right\} = H\left\{x\left(t-\frac{\pi}{2w_c}\right)\right\} $ where $H$ is Hilbert transform Accordind to my understanding of the following:

 


$$H\left\{X(f)\right\} = -j\;sgn(f)\; X(f)  \qquad(1)$$  

my work so far:

since $ x_h(t) = x\left(t-\frac{\pi}{2w_c}\right)$, Their Hilbert transforms are equal in time and frequency domains.Hence,

$$ H\left\{x_h(t)\right\} = H\left\{x\left(t-\frac{\pi}{2w_c}\right)\right\}, \:\text{where $H$ is Hilbert transform. 
} $$ 

Let  $ \: y(t) = x\left(t-\frac{\pi}{2w_c}\right)\ \: $.We have 

$$ H\left\{x_h(t)\right\} = H\left\{y(t)\right\} .$$ Therefore, 

$$H\left\{X_h(f)\right\} = H\left\{Y(f)\right\} \\ \\ $$ 

L.H.S

by applying what is stated in the picure, we have:
$$H\left\{X_h(f)\right\} = -j\;sgn(f)\;  X_h(f)   = -j\;sgn(f)\;  \left[-j \; X(f) \; sgn(f)\right] $$ 

after simplification:
$$\text {L.H.S} = -X(f)$$

R.H.S
$$H\left\{Y(f)\right\} = -j\;sgn(f)\; Y(f) = -j\;sgn(f)\; \left[X(f) \; e^{-j2\pi \left(\frac{f}{4f_c}\right)} \right ] = \text {R.H.S} $$  by simplifying $$ \frac{\pi}{2w_c} = \frac{\pi}{2\left[2\pi f_c\right]}  = \frac{1}{4fc} $$ and by the shifting property:

$$F \left\{x(t-t_0)\right\} = X(f) \; e^{-j 2\pi f t_0}  \\ \\ $$

$$ - X(f) \quad \ne  \quad -j\;sgn(f)\; \left[X(f) \; e^{-j2\pi \left(\frac{f}{4f_c}\right)} \right ] $$

Thank you in advance.
It is quite unclear what you're actually asking, and most of your equations have no meaning. But one thing that you need to understand is that the equation

$$\mathcal{H}\{x(t)\}=x\left(t-t_0\right),\qquad t_0=\frac{\pi}{2\omega_0}=\frac{1}{4f_0}\tag{1}$$

only makes sense for sinusoidal functions $x(t)$ (of frequency $f_0$). For non-sinusoidal signals $x(t)$ the Hilbert transform of $x(t)$ can only be written as an integral:

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau\tag{2}$$

where the improper integral is interpreted as the Cauchy principal value. In general, the Hilbert transform is not just a simple time shift.

For a sinusoidal signal $x(t)=\cos(2\pi f_0t+\phi)$ you can indeed show that the Hilbert transform corresponds to a delay of $t_0=1/(4f_0)$, i.e., in the frequency domain the following holds:

$$-j\,\textrm{sgn}(t)\,X(f)=X(f)e^{-j2\pi f t_0}=X(f)e^{-j\frac{\pi f}{2f_0}}\tag{3}$$