I came across the following, let $g : [0,1] \to \Bbb R$ be a concave function with $g(0) =0$ and $g(1)= \beta$. It implies $g(z) \geq \beta z$, $z \in [0,1]$. Why is the statement $g(z) \geq \beta z$ true?
$g : [0,1]\to\Bbb R$ is a concave function with $g(0) =0$ and $g(1)= \beta$. Show that $g(z) \geq \beta z$, $z \in [0,1]$.
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0This is pretty obvious if you graph it. – 2017-02-22
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0@Goldname Perhaps posting the graph would be helpful, then. – 2017-02-22
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0@TheCount Oh no, not you again. – 2017-02-22
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0@Goldname I don't know what you mean, but if graphing provides an obvious solution, it sounds like it would be helpful to post. – 2017-02-22
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1@Goldname how do you post a "generic" concave function? A continuous one gives the intution for sure, but I'm curious how you would explain how to extrapolate to the general case. – 2017-02-22
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0@AdamHughes Like you said, the secant line is below the graph of every concave function. So if you draw that line, which can be considered "least" concave, it becomes obvious. – 2017-02-22
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1@Goldname right, I agree, but I'm looking for where that isn't circular: the definitions are what imply the geometry, so I'm trying to see how you can get the "general shape" of a concave graph without already knowing it has this property and ending up using circular reasoning. i.e. "it's below the secant because I drew this graph where all the points are below the secant" – 2017-02-22
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0@AdamHughes This isn't circular reasoning. You want to draw a concave line, intuition says to consider the cases where the line approaches $\beta*z$ the most. That line is the the least concave line. It's not a rigorous proof but I'm sure you can extrapolate something out of it. – 2017-02-22
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1@Goldname how do you know what a concave graph looks like without the definitions though? Like, if I draw one squiggle and you draw another, how can we tell which (if any) are the thing we're calling "concave?" It doesn't seem you're advocating for using the graph as a proof, so I think what you might be trying to say is to use a graph of something you intuit as "concave" for motivation, in which case I agree, but I want to be sure since your original comment looks like "if you draw a picture, that's enough to answer the question," which it looks like you might not have intended to say. – 2017-02-22
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0@AdamHughes Why are you handicapping me from using the definition of concave? I never said you can draw a concave line while ignoring the definition of it. Also, I'm curious to know, are there rigorous proofs through visualization? I never heard of such things, but I definitely didn't mean that it was possible. – 2017-02-22
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1@Goldname the standard definition of concave is $f(x)$ is concave iff $-f(x)$ is convex, where convex means $f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(x)$ with $0\le\lambda\le 1$. I'm not saying you shouldn't use it, I was trying to figure out if you were using a words description or the actual inequality as it was unclear from what you wrote, and it's impossible to tell since there are a plenty of users who use intuitive definitions rather than rigorous ones. In short: I don't think you necessarily were talking in circles, but it wasn't clear from how you were explaining it. – 2017-02-22
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0@AdamHughes I wasn't aware of the formal definition, but I was aware of its existence. – 2017-02-22
2 Answers
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The secant line from $[0,1]$ is
$$y-0= {\beta -0\over 1-0}(x-0)$$
By definition of concavity, the secant line is below the graph of the function on any interval, so the result immediately follows.
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By definition of concavity on interval $[0,1]$, for any $\forall \lambda \in [0,1]\,$:
$$ g\big((1 - \lambda) \cdot 0 + \lambda \cdot 1\big) \;\ge\; (1-\lambda) \cdot g(0) + \lambda\cdot g(1) $$
Writing the above for $\lambda = z \in [0,1]\,$ with $\,g(0)=0\,$ and $\,g(1)=\beta\,$ gives $\,g(z) \ge z \cdot \beta\,$.