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Let $A$ be a countable and discrete subset of ${\Bbb R}^d$, $d \ge 2$, ofpositive upper Beurling density. Let $K$ denote a bounded symmetric convex setwith a smooth boundary and everywhere non-vanishing Gaussian curvature. It isknown that ${\mathcal E}(A)=\{e^{2 \pi i x \cdot a}\}_{a \in A}$ cannot serveas an orthogonal basis for $L^2(K)$ \cite{IKT01}. In this paper, we prove thateven approximate average orthogonality is an obstacle to the existence of anexponential frame in the following sense. Let $A$ be as above and $\phi \ge 0$be a continuous monotonically nonincreasing function on $[0, \infty)$ such thatthe approximate orthogonality condition holds \begin{align}\notag {\left(\frac{1}{2^j} \int_{2^j}^{2^{j+1}} \phi^p(t) dt \right)}^{1/p} \leq c_j2^{-j\frac{d+1}{2}} \quad \text{and} \quad |\widehat{\chi}_K(a-a')| \leq\phi(\rho^*(a-a')) \ \forall a \not=a , a,a' \in A, \end{align} where $\rho^*$is the Minkowski functional on $K^*$, the dual body of $K$. Then, if$$\limsup_{j \to \infty} c_j=0,$$ then the upper density of $A$ is equal to$0$, hence ${\mathcal E}(A)$ is not a frame for $L^2(K)$. The case $p=\infty$was previously established by the authors of this paper in \cite{IM2020}. Thepoint is that if ${\mathcal E}(A)$ is a frame for $L^2(K)$, then very few pairsof distinct exponentials $e^{2 \pi i x.a}, e^{2 \pi i x.a'}$ from ${\mathcalE}(A)$ come anywhere near being orthogonal. Our proof uses a generalization ofBourgain's result on pinned distances determined by sets of positive Lebesgueupper density in ${\Bbb R}^d$, $d \ge 2$. We also improve the $L^{\infty}$version of this result originally established in \cite{IM2020}. By using anextension of the combinatorial idea from \cite{IR03}, we prove that under the$L^{\infty}$ hypothesis, $A$ is finite if $d \not=1 \mod 4$. If $d=1$ mod $4,A$ may be infinite, but if it is, then it must be a subset of a line.

Approximate orthogonality, Bourgain's pinned distance theorem and exponential frames