Is the below filter linear phase
If

$$h(t)=\frac{1}{1+t^2}\tag{1}$$

denotes the filter's impulse response, then the filter is an infinite impulse response filter, simply because $h(t)$ has infinite support, i.e. it doesn't vanish anywhere (other than for $t\rightarrow \infty$). Note that if the definition $(1)$ of $h(t)$ is valid for all $t$, then the filter is not causal because $h(t)=0$ for $t&lt;0$ doesn't hold.

Since the impulse response is real-valued and it satisfies $h(t)=h(-t)$, its Fourier transform must be real-valued (and symmetrical). Since the Fourier transform is real-valued, the filter's phase is zero.

If you sample that impulse response using $5$ taps at times $t_{-2}=-2T$, $t_{-1}=-T$, $t_0=0$, $t_1=T$, $t_2=2T$, you get an FIR filter, but the zero-phase property remains unchanged due to the symmetry of the filter coefficients. For implementation, the filter impulse response needs to be shifted by two samples to make it causal. This causes a delay of two samples, which changes the filter's phase $\phi(\omega)$ from zero to a linear phase $\phi(\omega)=-2\omega$.