Showing a function is convex on $x^2+y^2\leq a^2$ 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....
 A: 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.
A: 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 $\phi$.
