Smartass/"Child-like" logic:
We have 11 identical coins, there can't be a fake if they're all identical, and if one is fake, then the rest are fake. The scale is not important. Therefore, 0 uses minimum.
Applied/Linear thinking potential solution:
My other answer is potentially 2 (but really 3), if you odd-man out, placing 10 on the scale (classic 2-panel scales), and you're holding the fake (under ideal circumstances) if it 50-50s, 5 on each side, as such scales tend to be used in such riddles, and hold the remaining coin to the side. The second time used is to confirm the fake in-hand (remove all but 1, and place the remaining #11 on the scale. If off-kilter, then you had the fake the entire time).
As for single-panel scales, that's a bit variable, but would be about 3 times, because you need to total sum of the weight, 2 coins to baseline at a constant weight for real coins (but becomes twice if something doesn't add up, and baselining again would confirm something's off, and you found your fake), and then with the 2 removed from the equation, apply the rest of the 9, and remove gradually until the discrepancy is removed, and you have a divisibility that lines up with the baseline value. About 3 times minimum is the closest answer, provided the 3rd measurement is a gradual progression of addition or removal, and technically still counts as the 3rd use.
Considering 1-Panel's method, you can still apply it to 2-Panel, and still only really need 3 uses to figure it out. So answer for both types of scales, 3 minimum. Just as well, if you decide to measure coins individually, 3 is still minimum. If under ideal circumstances, you have 2 same and 1 different. Odd-man-out would determine the "3rd" coin (under said ideal circumstances; out of the 3 measured) stands out as false.
Another method: 5-5-1. Measure 5, measure another 5, then measure the final 1. Sum/n (n=5 in this case) for the previous 2 stacks to find the odd man out, if both are equal, the #11 is the fake. If sum/n between them is not equal, then do the same method for the odd pile out, and repeat the process (2-2-1, and #5) until you find the fake. It's still 3 uses minimum required, especially under ideal circumstances; and the sub-divide is still 3 uses minimum; if still nothing, then 1-1-1 (T-T-F; again, another 3-min). So 5-5-1 (if one of the 5 have it) → 2-2-1 (if one of the 2 have it, keep one more)→ 1-1-1 → (T-T-F). 3 minimum uses of 3 minimum uses progressively; as long as you have an odd number, and a single discrepancy in that mix of constants. Identify by comparing the odd one out by dividing the sum by the # of coins every division, and compare it to the isolated lone coin.
So: (Sum1/x):(Sum2/x):y or y1:y2:y3; if not 1:1:1 or y(1:2:3)≠1, then work on the non-1.
Work with that, in that format, until you whittle down to the final 3 coins, which still looks the same, and you'll find your fake eventually. 7-9 measurements maximum using this method, given 11 coins.
Lateral thinking at it's finest, answer:
Alternatively, 0 is also an answer, provided you test coins like you're checking if an egg is raw or hard-boiled by spinning them. Just as well, if you have good enough ears, you can bang the coins together, or drop them and listen for anything odd, to determine the fake; thus the scale doesn't even need to be used at all, therefore: 0.
Once is one more answer, if you feel like throwing the coins at the scale, and listening for anything odd. Technically, the scale is still being used.
Overall, how I see it, Get the sum (determine the fake is there), Make a baseline/control sample (to get a constant weight amongst constant identicals), And if anything was odd with either of the previous steps, try 2 new coins to confirm baseline/control sample to ensure the previous was not containing the fake (or by luck, you got the fake in this run). Thus 3 uses minimum. 7 uses maximum is another answer for that too, if you want, for the 2-panel. 50-50 2 coins at a time; by the time you find one that doesn't fit in, remove one side. If 50-50, you got the fake, if not, then the coin removed wasn't fake, and the coin left behind is fake.
Being the lazy oddball type, I still prefer the 0-use answers. They're more fun.