Prove the concavity of $F(x,y) =\ln(x)+y$ by arguing the definition of concavity.

Question

I need help with this problem.

Prove the concavity of $F(x,y) = \ln(x) + y$ by arguing the definition of concavity.

A function $f$ is concave is for any $x_0, x_1 \in \mathbb{R}^2$ and $t \in [0,1]$, $$ f((1 - t) x_0 + t x_1) \geq (1 - t) f(x_0) + t f(x_1) $$

Also, I am not allowed to use the theorem that the sum of concave functions is concave.

Do you already have the result that the log function is concave? — Brian M. Scott, Apr 08 '13 at 23:39

Brian M. Scott · Accepted Answer · 2013-04-09T00:31:43.560

3

You need to show that for any $\langle x_0,y_0\rangle$ and $\langle x_1,y_1\rangle$ and any $t\in[0,1]$,

$$\ln\big(tx_0+(1-t)x_1\big)+ty_0+(1-t)y_1\ge t(\ln x_0+y_0)+(1-t)(\ln x_1+y_1)\;;$$

after a little simplification this becomes

$$\ln\big(tx_0+(1-t)x_1\big)\ge t\ln x_0+(1-t)\ln x_1\;.$$

If you already know that the log is concave, you’re done at this point.

edited Apr 09 '13 at 00:31

answered Apr 08 '13 at 23:59

Brian M. Scott

What if I do not know that the log is concave, how can I prove this inequality? – PooperScooper Apr 09 '13 at 00:03
1

@Nik: Use the second derivative. – Brian M. Scott Apr 09 '13 at 00:05
Is there a way of proving it without using the derivative? That is another part of the question. – PooperScooper Apr 09 '13 at 00:29
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@Nik: I’ve not really thought about it, but I shouldn’t be at all surprised: the inequality is equivalent (after exponentiation) to $t_x_0+(1-t)x_1\ge x_0^tx_1^{1-t}$, which is a special case of the weighted AM-GM inequality, whose unweighted case has many proofs. – Brian M. Scott Apr 09 '13 at 00:35

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