 Solution

Use the canonical chiral-superfield component expansion $\Phi =\varphi +\sqrt{2}\theta \psi +\theta \theta F$ in chiral coordinates, with $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$. Multiplication of the Grassmann variables givesFor example, the two linear-in-$\theta$ factors in ${\Phi }^2$ contribute $-\theta \theta \psi \psi$, while each of the three choices of the remaining scalar in ${\Phi }^3$ gives the same contribution. Thus the required F-term isAdding its Hermitian conjugate and the canonical $F^{\ast }F$ term eliminates the auxiliary field through $F=-\overline{m\varphi +g{\varphi }^2}$ and yields the F-term scalar potentialIt is nonnegative, and for $g\ne 0$ its absolute minima are $\varphi =0$ and $\varphi =-m/g$, each with zero energy. They coincide when $m=0$. If $g=0,m\ne 0$, only $\varphi =0$ is a minimum; if $g=m=0$, every $\varphi$ is a minimum. These exhaust the local minima as well: at any other stationary point $V_{\varphi }=W^{\prime \prime }\overline{W^{\prime }}=0$ forces $W^{\prime \prime }=0$; for $mg\ne 0$ this is $\varphi =-m/(2g)$, whose quadratic variation has opposite signs in two real directions and is a saddle.

Expand around $\varphi =0$, writing $\varphi =h=(h_1+i{h}_2)/\sqrt{2}$. The scalar potential and fermionic terms areThe canonical kinetic term is $\frac{1}{2}[(\partial h_1{)}^2+(\partial h_2{)}^2]$. Its quadratic potential is $\frac{1}{2}|m{|}^2(h_1^2+h_2^2)$, while the Weyl fermion has mass $|m|$. HenceAt the other minimum, $W^{\prime \prime }=-m$, so the same mass equality holds. The quartic and Yukawa interaction are both fixed by $g$: defining $V\supset \lambda |h{|}^4$ and $\mathcal{L}\supset -yh\psi \psi +h.c.$ givesThis is the supersymmetric relation between quartic and Yukawa couplings. The printed equality of the two couplings refers to their common supersymmetric origin; literal equality of the conventional coefficients $\lambda$ and $y$ is not true for arbitrary $g$. If instead the Yukawa term is defined as $-\frac{1}{2}yh\psi \psi$, then $y=2g$ and $\lambda =|y{|}^2/4$.

To establish perturbative superpotential non-renormalization, use a Wilsonian effective action with a fixed positive infrared cutoff, so integrating out modes does not introduce massless infrared singularities. Promote $m,g$ to nondynamical chiral superfields, or spurions. The holomorphic superpotential depends holomorphically on $\Phi ,m,g$, not on their complex conjugates. Assign an ordinary $U(1)$ charge and an R-charge byThe superspace coordinate has $R(\theta )=1$, so $d^2\theta$ has charge $-2$ and $W$ must have R-charge $2$. Both original monomials satisfy the two selection rules. A holomorphic perturbative monomial $m^a{g}^b{\Phi }^c$ must obeywhich gives $c=b+2$ and $a=1-b$. Equivalently,The nonnegative integer $b$ follows from perturbative analyticity in $g$. Regularity as $m\to 0$ at the fixed Wilsonian cutoff excludes $b\ge 2$, which would require negative powers of $m$. The remaining quadratic coefficient is fixed to $c_0=1/2$ by the free theory at $g=0$. The coefficient $c_1$ is $1/3$: a three-point contribution at first order in $g$ has just one cubic vertex, uses all three legs externally, and cannot contain a loop. A loop correction would require additional interactions, whereas the selection rules and holomorphy have already excluded the corresponding monomials. ThereforeThis proof concerns the local Wilsonian superpotential. The Kähler potential can acquire a wave-function factor $Z$, and canonical normalization changes the couplings to $m_c=m/Z$ and $g_c=g/Z^{3/2}$. Such running, and nonlocal infrared terms in a one-particle-irreducible action, do not contradict the holomorphic result.