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QUESTION F(x) =-x for x>=0 and F(x)=x for x<=0 Is the function convex/(strictly), concave/(strictly)

I have attempted the answer but got strictly concave but isnt a discontinuous function meant to be neither convex nor concave?

Thanks for you help in advance :)

Attempt at answer

erin
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  • $f(x)$ is continuous! – JCW Oct 28 '14 at 23:21
  • So is my analysis correct then? – erin Oct 28 '14 at 23:22
  • I thought it was continuous because it was broken into 2 different sections. Whoops my mistake – erin Oct 28 '14 at 23:23
  • Not quite. You can't just consider individual values, but must show that your claimed property holds for all $ x, y$. Can you see why it definitely isn't strictly concave? (consider, say, $ x, y > 0$). – JCW Oct 28 '14 at 23:28
  • Yes that makes sense because if you let x=1, y=3, lamda=.5 you get that they are equal. – erin Oct 28 '14 at 23:36
  • so can you say that the function is concave but not strictly concave as either >= ? – erin Oct 28 '14 at 23:37
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    You could, but you still need to do a fair bit of work before you can justify that answer. You couldn't ask for a nicer function! There are three cases: (i) $ x,y > 0 $, (ii) $x > 0, y < 0$, and (iii) $x,y < 0$. The first and third are immediate, because $f$ is nice (just write down the condition for concavity, evaluating $f$ at $x, y$, and you'll see). The second is trickier but not too difficult. Let me know how you get on and I'll write up a proper answer if required. – JCW Oct 28 '14 at 23:51
  • I posted my attempt below. thanks – erin Oct 29 '14 at 00:21

2 Answers2

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What you have is a piecewise function, but in this case, it is continuous.

The idea of strict concavity is that, for any two points on the graph, the segment connecting the two points lies below the graph except at those two points. For concavity, this segment must simply lie below or on the graph at all points.

You should be able to find two points whose connecting segment meets the graph in between the endpoints, so this function is not strictly concave. However, it is, indeed, concave.

To show this, you'll need to proceed by cases, depending on the signs of $x$ and $y.$ Note that you should not specify values for $x$ or $y,$ nor should you specify a value for $\lambda.$ See what you can do, and let me know if you're still stuck.

Cameron Buie
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I tried to do it generically for all 3 cases but got a bit stuck. Here is my attempt

Not sure if what I have done is right. and I am a bit confused what the outcome would be for case 2

erin
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  • You're on the right track. For case 2, you may need to split that case up further, depending on the value of $\lambda.$ Also, don't forget the possibility that $x=0$ or $y=0.$ – Cameron Buie Oct 29 '14 at 00:28
  • Thank you for your help :) – erin Oct 29 '14 at 00:32
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    You're very welcome. As a further hint, the kicker for that is to figure out which $\lambda\in[0,1]$ make $x+(1-\lambda)y<0$ and which make $x+(1-\lambda)y\ge0.$ – Cameron Buie Oct 29 '14 at 00:33