The acceleration of the center of mass for a rolling object down an incline is given by:

$$ a = \frac{g \sin\theta}{1 + \frac{I}{m r^2}} $$

For a uniform solid cylinder, the moment of inertia about its central axis is:

$$ I = \frac{1}{2} m r^2 $$

Substituting this into the expression for acceleration:

$$ a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta $$

For an incline of $$45^\circ$$, we have:

$$ \sin 45^\circ = \frac{\sqrt{2}}{2} $$

Thus, the acceleration becomes:

$$ a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3} $$

Therefore, the linear acceleration of the cylinder's axis is:

$$ \frac{\sqrt{2} g}{3} $$