Show that $g$ is a low pass filter

Question

Let $x(t)$ be a signal.

Let $$g(t) = \int_{-\infty}^{+\infty}x(t-\varepsilon)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{\varepsilon^2}{2\sigma^2}}\mathrm{d}\varepsilon$$

I would like to show that $g(t)$ is a low pass filter for the signal $x$.

Answer 1

As mentioned in the comments, $g(t)$ is the convolution of your signal with some function (the impulse response of your filter) which is a Gaussian waaveform. That is, $g(t)$ is the output of your filter. You just need to verify that a Gaussian waveform in frequency domain decays as $f\to \infty$ to confirm that your filter is indeed lowpass. This is true as the Fourier transform of a Gaussian is a Gaussian waveform.