= Solution
A spherical three-<momentum> <ultraviolet cutoff> gives
$$
\mathcal E(\Lambda)=\frac1{4\pi^2}\int_0^\Lambda dp\,p^2\sqrt{p^2+m^2}
=\boxed{\frac{\Lambda\sqrt{\Lambda^2+m^2}(2\Lambda^2+m^2)-m^4\operatorname{arsinh}(\Lambda/m)}{32\pi^2}}.
$$
For $m>0$ and $\Lambda\gg m$, the <asymptotic expansion> is
$$
\boxed{\mathcal E(\Lambda)=\frac{\Lambda^4}{16\pi^2}+\frac{m^2\Lambda^2}{16\pi^2}
-\frac{m^4}{32\pi^2}\log\frac{2\Lambda}{m}+\frac{m^4}{128\pi^2}+O(m^6/\Lambda^2).}
$$
In particular, the leading <ultraviolet divergence> is quartic. For $m=0$, $\mathcal E=\Lambda^4/(16\pi^2)$ exactly.
Write $\varphi=\langle\psi\rangle$. Under the proposed identification, the <scalar potential from a field-dependent momentum cutoff> is $V_{\mathrm{ind}}(\varphi)=\mathcal E(|\varphi|)$. It is even and nonnegative, and its derivative with respect to $|\varphi|$ is $|\varphi|^2\sqrt{\varphi^2+m^2}/(4\pi^2)>0$ away from zero. Consequently its only <global minimum> is at $\varphi=0$: this contribution alone does not cause <spontaneous symmetry breaking>. At large $|\varphi|$ it grows quartically. At small $|\varphi|$ with fixed positive $m$, however,
$$
V_{\mathrm{ind}}(\varphi)=\frac{m|\varphi|^3}{12\pi^2}+O(|\varphi|^5/m).
$$
Thus the quartic sketch is an asymptotic shape, not the exact small-field behavior for a massive original <field>. Identifying a regulator with a new <field> is an additional modeling prescription; an actual <effective potential> also depends on the new <field>'s dynamics and any other allowed terms.
\Image[/past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2018-cutoff-induced-potential.png]
{title=<Scalar potential> obtained by identifying a <ultraviolet cutoff> with a <scalar field> expectation value}
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