If a variable frequency ac source is connected to a capacitor then with decrease in frequency the displacement current will:

decrease

Current through capacitor, $I=\frac{E}{X_{C}}=\frac{E}{\frac{1}{\omega C}}=\varepsilon CE=2 \pi vCE$ or $I \propto v$

∴ decrease in frequency v of ac source decreases the conduction current. As displacement current is equal to conduction current, decrease in v decreases displacement current in circuit.