After thinking about Smith charts I'd like to change my answer here. First thing is test to see whether I can get an image in here. I guess I can so, OK here goes. You can't really convert the antenna itself but you can 'match' it to the transmitter/receiver. Assuming you had a network analyzer available to scan a 5.8 antenna at 2.4 GHz the antenna would appear on a Smith chart display somewhere like where the blue dot is on the attached picture. This is a terrible match with the antenna radiation resistance being about (multiply all numbers by 50) .09 (you want 1) with a reactive component of about -0.2 (you want 0). This results in a VSWR of 12.7:1 (terrible) and a return loss of -1.37 db. Most of the power delivered to this antenna will be reflected back to the transmitter.
To match move the blue dot along the dashed line circle (counterclock wise) until you hit the green circle. This is easily done by connecting a short (about 0.15 wavelength) piece of transmission line (coax) to the antenna. The green circle is the locus of points with conductance equal 1 (G = 1) so the conductance looking in to the other end of that coax is G = 1. The blue dot, before and after moving it represents impedance and the reactive impedance of the moved dot is about -2.8j (it is close to the -0.3 line and the j denotes it as a reactance). We could cancel it by putting a reactance of + 2.8j in series with the short piece of cable. A capacitor of 1/(2*pi*2.4E-3*2.8*50) = 0.47pf would do but that's pretty small so we look at another approach and that is to convert the impedance at the shifted point to the admittance at that point and this is done by reflecting the shifted point along the straight line through the center of the diagram until the heavy red unit circle is met on the other side. The admittance at this point is Y = 1 + 3.5j (1 because the red circle is the G=1 circle and 3.5 is obtained by eyeballing the distance along the arc between the 3 and 5 curves. If we add a conductance G' = 0 - 3.5 j in parallel at the input to the short cable end (effectively moving along the G=1 circle towards the origin we get G'' = 1 + 0j, the intersection of the heavy red line and heavy red circle. An easy and convenient way to obtain a conductance of 0 - 3.5j is by taking another piece of transmission line shorted at one end. To see how long this should be we plot -3.5j on the outer circle of the diagram and project that thru the center to see where it hits the other side or just calculate 1/-3.5j = 0.29j (j is a rather special number such that 1/j = -j) which says that j*j = -1) and plot that on the outer circle where the pencil dot is in the picture. Starting at where the heavy red lone intersects this outer circle (9 o'clock) and rotating clockwise until the pencil dot is reached you will go 9% of the way around the circle. Since once around the circle corresponds to two wavelengths we find that we need 4.5% of a wavelength of cable to put in parallel with the first piece of cable. This is what devices like the 'tee' pictured in #11 are for. The input to the third leg on the tee is G = 1 + j0 and the reciprocal of that is the impedance Z = 1 + j0. Un-normalizing by multiplying by 50 we have Zs = 50 + j0 Ω - a perfect match.
So conceptually you could build a matching network for that 5 GHz antenna and get it on the air. It probably wouldn't be practical to do it with coax as the pieces would be so short but it could be done with strip line. The real purpose of this is to illustrate the ease with which matching problems can be solved if one has the Smith chart to look at.