= Solution
For positive integer $k$, the equivalence is the <bounded invariant norm characterization of cusp forms>. The three conditions below work together: meromorphicity and the bound remove interior poles; the transformation law produces a periodic local function at every <modular cusp>; the same bound makes that function tend to zero at the <modular cusp>. Conversely a <cusp form> has exponentially decaying <Fourier expansions of a modular form>, and finitely many compact core regions control its invariant <norm> everywhere. These implications are proved under the corresponding labelled items.
\b[Without a positive-weight convention, weight zero is a genuine exception.] The function $f=1$ satisfies all three printed conditions at $k=0$ for every <congruence subgroup>, but is not a <cusp form>. Indeed a weight-zero <cusp form> would descend to a <holomorphic function> on the compact curve $X(\Gamma)$ and hence be constant, and vanishing at a <modular cusp> would make it zero. Thus the literal unrestricted claim needs the qualification $k>0$.
If negative integer weights are included, both sides contain only the zero function. To check the bounded side, each <modular cusp> translate has at most power growth in $y$, hence at most a power of $\log(1/|q_h|)$. For a negative <Laurent coefficient> of order $m$, Cauchy's coefficient estimate bounds its absolute value by $C r^m(\log(1/r))^{|k|/2}\to0$, so there is no negative coefficient and every <modular cusp> is <holomorphic>. If such a nonzero negative-weight form existed, the product of its slash translates over $\Gamma\backslash SL_2(\mathbb Z)$ would be a nonzero <holomorphic> level-one <modular form> of negative weight. Right multiplication permutes the translates, and all <modular cusp> factors are bounded, so that product really is a level-one form. Its existence contradicts the valence formula. This also excludes negative-weight <modular cusp> forms. The nonzero counterexample is specifically weight zero.
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