Google Earth Engine pan sharpening of Landsat 7 image is not producing the correct true color image

Question

I'm currently trying to produce a true color pansharpened Landsat 7 image in Google Earth Engine. I've included my code below and a subset of the resulting image. I'm simply using the rgbToHsv function to convert the true color bands for Landsat 7 (B3, B2, B1) to HSV then subbing in the panchromatic band and converting back to RGB. This process has worked for Landsat 8 and even worked on this same Landsat 7 image when I included the false-color bands (B4, B3, B2). However code below image results in a sharpened image but for some reason vegetation appears cyan-blue instead of the expected dark green color. I've included a screenshot of the resulting image. The issue with the color is most noticeable in active agriculture plots. I can't figure out why the resulting image is blue in places. Initially I thought I included the wrong bands for True Color with Landsat 7 but that doesn't seem to be the issue.

Does it have something to do with the rgbToHsv or hsvToRgb functions?

var L7image = ee.Image('LANDSAT/LE07/C01/T1_TOA/LE07_037037_20000606').clip(sa);
var L7image_hsv_truecolor = L7image.select(['B3','B2','B1']).rgbToHsv();
var L7image_sharp_truecolor = ee.Image.cat([L72000_hsv_truecolor.select('hue'),L72000_hsv_truecolor.select('saturation'),sngimgL72000_037037.select('B8')]).hsvToRgb();

Answer 1

Your code is correct, but this does not work for Landsat 7 as its panchromatic band includes a range of wavelengths that is both wider and shifted longer (less blue, more infrared) than visible light.

This differs from Landsat 8 which has a panchromatic band much closer to the visible bands.

You can still pansharpen Landsat 7, but the techniques to do so are more involved than HSV swapping of luminance.