Showing a function is convex on $x^2+y^2\leq a^2$

Question

This is a question from my assignment about which I have no idea:
Let $f(x,y)=\phi(x^2+y^2)$,where $\phi$ is of class $C^2$ ,increasing and concave.

Show that $f$ is convex on disk $x^2+y^2\leq a^2$ if and only if $\phi'(u)+2u\phi''(u)\geq 0$ whenever $0\leq u\leq a^2.$

Can anyone please help me with this....

where do you have problems? Start with the definition of convexity for $f$.. where does that takes you? — Ant, Oct 15 '14 at 13:07

score 2 · Answer 1 · answered Oct 15 '14 at 13:12

Hint: Consider $(x_1,y_1), (x_2,y_2)$ such as $$ x_1^2 + y_1^2\le a^2\\ x_2^2 + y_2^2\le a^2 $$ Consider $g:[0,1]\to \Bbb R$, $g(t) = f(t (x_1,y_1) + (1-t) (x_2,y_2))$.

$f$ is convex iff for every choice of $(x_1,y_1), (x_2,y_2)$ the function $g$ is convex.

hickslebummbumm · Accepted Answer · 2014-10-16T22:29:40.280

2

Warning: It has been pointed out to me that the proof is flawed as convexity over $r$ does not necessarily imply convexity over $(x,y)$. Any suggestions as to how to repair it are greatly appreciated.

You can write this in polar coordinates

\begin{align*} f(x,y) \rightarrow f(r) = \phi(r^2) \end{align*}

A twice differentiable function (of one variable) is convex if and only if $\frac{\mathrm{d}^2}{\mathrm{d}r^2} f(r) \geq 0$ on the entire domain. We have

\begin{align*} &\frac{\mathrm{d}}{\mathrm{d} r} f(r) = 2r \phi'(r^2) \\ & \frac{\mathrm{d}^2}{\mathrm{d}r^2} f(r) = 2 \phi'(r^2) + 4r^2 \phi''(r^2) \end{align*}

Now, $2 \phi'(r^2) + 4r \phi''(r^2) \geq 0 \implies \phi'(r^2) + 2r^2 \phi''(r^2) \geq 0$. Now introduce $u = r^2$. The condition reduces to

\begin{align*} \phi'(u) + 2u \phi''(u) \geq 0 \end{align*}

As $\phi$ is increasing, we have $\phi' \geq 0$ but as $\phi$ is concave, too, we have $\phi'' \leq 0$, therefore, this condition has to hold and depends on the properties of $\phi$.

edited Oct 16 '14 at 22:29

answered Oct 15 '14 at 13:16

hickslebummbumm

1,451

and why is "A twice differentiable function (of one variable) is convex if and only if d^2/dr^2 f(r)≥0 "? – mookid Oct 15 '14 at 13:26
because then $f'(r)$ is monotonically non-decreasing, i.e. the derivative only increases (or stays constant). Draw such a function and you shall see that it is indeed convex. Alternatively you may google for this very well known proof. – hickslebummbumm Oct 15 '14 at 13:30
I get that. My question is how to pass from convex as a function of the variable $r$ to convex as a function of $(x,y)$. – mookid Oct 15 '14 at 13:32
You're right, this proof is flawed, any suggestions as to how to repair it ? – hickslebummbumm Oct 16 '14 at 22:11
well, check my answer. Too bad this one has been accepted. – mookid Oct 16 '14 at 22:12
I didn't upvote it – hickslebummbumm Oct 16 '14 at 22:14
sure, as it is your answer ;) start with an upvote of mine :p – mookid Oct 16 '14 at 22:18
there you go :-) – hickslebummbumm Oct 16 '14 at 22:19
that is nice! now the problem is that the accepted answer is not good. Maybe you can start with an edit of your post. I don't know if there is a patch to fix the problem and still use your method. – mookid Oct 16 '14 at 22:23
How should I edit my post if I don't know how to correct it ? – hickslebummbumm Oct 16 '14 at 22:25
Just add a warning at the top of it, saying that the answer is not quite right at the moment, so that anyone who read it later does not make the same mistake. – mookid Oct 16 '14 at 22:26

Showing a function is convex on $x^2+y^2\leq a^2$

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