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Suppose the impulse response of a discrete space filter is $h(m, n)$. If $h(m, n)$ is a low pass filter, then $h(m, n)\ge 0$ for $m, n \in \mathbb{N}$.

Is the above statement true or false? If true, then prove it. If false, then provide an example of a discrete space low pass filter for which $h(m, n) \lt 0$ at some locations.

By natural intuition, I feel that even if the Gaussian kernel in spatial domain has negative coefficients or the averaging filter has negative coefficients, both of them would still be a low pass filter.

But how to prove it for the general case?

I tried to apply the definition of ideal low pass filter and then take an inverse DFT but did not arrive at any general expression. Some insight on this would really help.

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    $\begingroup$ Are you just asking if you were at a location given by m=0, n=0, where m represents left / right and n represents forward and backward, to prove that you can move to the left just as much as you can move to the right, or backward just as much as you can move forward? We can't do this with time in our causal world, but we can certainly do this in space. Still a non-causal low pass filter in time is still a low pass filter. Just take the Fourier Transform of its impulse response o get the frequency response and you prove that. Do the same thing with space if you need to. $\endgroup$
    – Dan Boschen
    Commented Nov 2 at 7:49
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    $\begingroup$ You just need a counterexample to disprove the claim: Here it is! $\endgroup$
    – Matt L.
    Commented Nov 2 at 9:58
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    $\begingroup$ This one seems trivial to disprove: the impulse response of an ideal lowpass filter is a sinc() function $\sin(x)/x$ which has plenty of negative coefficients. $\endgroup$
    – Hilmar
    Commented Nov 2 at 11:58
  • $\begingroup$ Thanks everyone. It makes sense now. $\endgroup$
    – Curiosity
    Commented Nov 7 at 18:24

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