 Solution

A K correction accounts for observing a redshifted spectrum through a fixed photometric passband, rather than the corresponding emitted band. In the same R-band astronomical magnitude convention,The luminosity distance already incorporates bolometric dimming; the additional correction depends on spectral shape and band response. At $z=3$, observed R-band light at $660\mathrm{nm}$ was emitted at $165\mathrm{nm}$, in the far ultraviolet. It is not a direct measurement of rest-frame R-band luminosity.

To fix the sign, let $L_{\nu }$ be emitted spectral luminosity and approximate R as narrow. Counting photon energy, arrival time and the transformed frequency interval gives $F_{{\nu }_o}=(1+z)L_{(1+z){\nu }_o}/(4\pi d_L^2)$. ThusFor an ordinary old elliptical galaxy, the far-ultraviolet spectrum is weak relative to the optical, so typically $L_{\nu }(165\mathrm{nm})/L_{\nu }(660\mathrm{nm})<1/4$. Then $K_R>0$: neglecting it overestimates the numerical absolute magnitude and underestimates the R-band luminosity. For a young, unobscured star-forming galaxy, hot stars make the far-ultraviolet strong, with an approximately flat $L_{\nu }$ spectrum often serving as a useful model. It gives $K_R=-2.5{\log }_{10}4\simeq -1.51$: neglecting it underestimates the numerical absolute magnitude and overestimates the R-band luminosity. Strong dust extinction or an unusual spectrum can reverse this latter sign; a galaxy label alone does not prove an exact correction. The band-integrated convention is given in Hogg and collaborators' definition of the K-correction.