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10393370 No.10393370 [Reply] [Original]

I'm struggling with writing this proof mathematically (intuitively it is easy). Can anyone help me with the following question regarding quasiconcavity? I would greatly appreciate it.

>> No.10393646

For all x^1, x^2 in D and t in [0,1], h(x) is either f(x) or g(x). Whatever the case, quasiconcavity thus gives that h(x) is superior to the minimum of the values of either f or g at x^1 and x^2. Since h is the most chibi of the three functions, this means that h(x) is >= h(x^1) and h(x^2). Therefore h is quasiconcave.

>> No.10393652

sorry, meant "minimum of h(x^1) and h(x^2)"

>> No.10393843

>>10393646
yeah I thought the same, but is writing that down sufficient as a proof?

>> No.10393857
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10393857

>>10393370
[math] min[f(x^t), g(x^t)] \geq min[min[f(x^1), f(x^2)], min[g(x^1), g(x^2)]] [/math]