There is the size hierarchy theorem for non-uniform circuits.

Do we have any size hierarchy theorem for any kind of uniform circuits ?

(By uniform here, I mean DLOGTIME uniform. But I don't know if this matters.)

For example, do $O(n)$-size constant depth threshold circuits have less power than the ones with size $O(n^{10})$ or even super-polynomial ?

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    $\begingroup$ The usual diagonalization argument works. $\endgroup$ – Kaveh Aug 14 '13 at 15:24
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    $\begingroup$ @Kaveh, could you explain what you mean by this? Thanks $\endgroup$ – Igor Shinkar Aug 19 '13 at 4:24
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    $\begingroup$ A slight variant of this question has been asked at cstheory.stackexchange.com/questions/5110/… . Also see a similar question about circuit depth at cstheory.stackexchange.com/questions/12872/… . $\endgroup$ – argentpepper Aug 20 '13 at 17:34
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    $\begingroup$ @Kaveh, The usual diagonalization argument does not answer (the standard interpretation of) his last question, because you need a function of a fixed depth which cannot be computed by circuits of $O(n)$ size and any depth. $\endgroup$ – Manu Dec 17 '13 at 20:59
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    $\begingroup$ @Kaveh, it's not clear to me that a simple diagonalization works even for unbounded depth circuits, if you insist on logtime uniformity. $\endgroup$ – Manu Dec 18 '13 at 15:10

Regarding your last question: The paper Size-Depth Trade-offs for Threshold Circuits shows that the parity function requires depth-$d$ threshold circuits with $\ge n^{1+\epsilon(d)}$ wires, which is tight up to the function $\epsilon$. But for gates not even $\Omega(n)$ lower bounds are known.


Not sure about what kind of results you seek but here what I know for sub-classes of $AC^0$ (constant depth and polynomial size Boolean circuits):

The separation between $AC^0$ and its linear fragment (namely $LAC^0$) is known since 96. It is a result of Chaudhuri and Radhakrishnan : "Deterministic restrictions in circuit complexity". This result seems to be non-uniform.

I heard about separation between each layer $n^k$ but unfortunately I don't know any ref for that.


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