Prove that $\phi :(x,y) \mapsto x^{\alpha} y^{1-\alpha}$ is concave

Question

Let $G=[0,\infty) \times [0,\infty)$, $\alpha \in (0,1)$ and $$\phi (x,y)=x^{\alpha} y^{1-\alpha}$$ Then $\phi$ is concave; that is, $-\phi$ is convex.

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This is left as an exercise in the book that I'm currently reading,and I think I have found a proof,which is shown below:

Fix $x,y \in [0,\infty)$ and $n \in \mathbb N$, where $\mathbb N$ always denotes the set of all positive integers,then we define $\alpha :=k/n$ for an arbitrary positive integer no greater than n so that $k/n \le 1$.

Let $\lambda \in (0,1)$,and we define

$$z_k:=(\lambda x + (1-\lambda)y)^(k/n)$$

and

$$w_k:=(\lambda x)^(k/n)+((1-\lambda)y)^(k/n)$$

then it follows that

$$((z_k)^n)^(1/k)=\lambda x+(1-\lambda)y$$

and that

$((w_k)^n)^(1/k) =[(\lambda x)^(k/n)+((1-\lambda)y)^(k/n)]^(n/k)$

$={\lambda x}[1+((1-\lambda)y/{\lambda x})^(k/n)]^(n/k)$

$\ge {\lambda x}[1+((1-\lambda)y/{\lambda x})^(k/n \cdot n/k)]$

$={\lambda x}[1+((1-\lambda)y/{\lambda x})]$

$=\lambda x+(1-\lambda)y$

$=((z_k)^n)^(1/k) $,

since it can be shown that $(1+x)^{\alpha} \ge 1+x^{\alpha}$ whenever $x \in [0,\infty)$ and $\alpha \in [1,\infty)$.

Therefore,$w_k \ge z_k$ holds,and hence the continuity of $\phi (x,y)$ with respect to $\alpha$ shows that

$(\lambda x+(1-\lambda)y)^{\alpha} \le (\lambda x)^{\alpha}+((1-\lambda)y)^{\alpha}$ for all $\alpha \in (0,1)$.

Finally,let $\lambda \in (0,1)$,and let $u:=(x_1,y_1),v:=(x_2,y_2) \in G$,then we obtain

$\phi (\lambda u+(1-\lambda)v)$

$=(\lambda x_1+(1-\lambda)x_2)^{\alpha}(\lambda y_1+(1-\lambda)y_2)^{1-\alpha}$

$\ge ((\lambda x_1)^{\alpha}+((1-\lambda)x_2) ^{\alpha})((\lambda y_1)^{1-\alpha}+((1-\lambda)y_2)^{1-\alpha})$

$\ge {\lambda x_1}^{\alpha} {\lambda y_1}^{1-\lambda}+{(1-\lambda)x_2}^{\alpha}+{(1-\lambda)y_2}^{1-\lambda}$

$=\phi (\lambda u)+\phi ((1-\lambda)v) $,

proving that $\phi(x,y)$ is concave.

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Is there anything wrong with my proof?I agree that my proof seems a little bit clumsy,so can anyone show me a succinct proof?Thank you!

Answer 1

An alternative is to look at Hessian of $\phi$ and see that it is negative definite.

Answer 2

I haven't read your proof in detail. I gave it a quick glance and my main feedback is that people would be able to read your proof more easily if you explained your thought process, ideas and motivations before doing calculations. E.g. Why did you set $\alpha = k/n$ ? A lot of what you wrote would be clearly if written in reversed order (which is probably the way you originally came to your proof). Both styles of writing such proofs are common but your style demands more mental effort from the reader.

Anyway, some general tips for approaching such questions of deducing e.g. certain functions are convex/measurable/differentiable, certain sets are open/closed, certain algebraic structures are isomorphic... etc. Try to do it in this order:

1. Keep a mental cache of the most fundamental and commonly occurring examples of objects with the property you are looking for. Understand why they are good examples and/or occur frequently.

2. Know the "algebra" of that property i.e. what rules apply in building such objects from other, simpler objects of that type. Results such as: Sum of two convex functions is convex, composition of a convex function with a convex function is convex, supremums of arbitrary sets of convex functions is a convex function, etc.

3. Know the basic useful lemmas/theorems for deducing/disproving the property. Results such as the First Order condition for convexity: If $f$ has a convex domain and is differentiable, then $f$ is convex if and only if $f(y) \geq f(x) + \nabla f(x) \cdot (y-x)$ for all $x,y$ in the domain of $f,$ i.e. the graph of the function lies above every tangent. Or the second order condition for convexity (that the Hessian evaluated at any point in the domain is positive semi-definite.

4. Only as a last resort, try to verify the definition directly.