Let \(0 \leq s \leq 1\), and let \(\mathbb{P} := \{(t,t^{2}) \in \mathbb{R}^{2} \colon t \in [-1,1]\}\). If \(K \subset \mathbb{P}\) is a closed set with \(\operatorname{dim}_{\mathrm{H}} K = s\), it is not hard to see that \(\operatorname{dim}_{\mathrm{H}} (K + K) \geq 2s\). The main corollary of the paper states that if \(0 < s < 1\), then adding \(K\) once more makes the sum slightly larger:
\( \operatorname{dim}_{\mathrm{H}} (K + K + K) \geq 2s + \epsilon,\)where \(\epsilon = \epsilon(s) > 0\). This information is deduced from an \(L^{6}\) bound for the Fourier transforms of Frostman measures on \(\mathbb{P}\). If \(0 < s < 1\), and \(\mu\) is a Borel measure on \(\mathbb{P}\) satisfying \(\mu(B(x,r)) \leq r^{s}\) for all \(x \in \mathbb{P}\) and \(r > 0\), then there exists \(\epsilon = \epsilon(s) > 0\) such that
\(\|\hat{\mu}\|_{L^{6}(B(R))}^{6} \leq R^{2 - (2s + \epsilon)}\)
for all sufficiently large \(R \geq 1\). The proof is based on a reduction to a \(\delta\)-discretised point-circle incidence problem, and eventually to the \((s,2s)\)-Furstenberg set problem.