How to prove that $er^2 \leq e^{r^2}$ for all $r \in \mathbb{R}$

Question

I'm trying to prove the function $f:\mathbb{R}^2 \to \mathbb{R}$ defined as $$f(x,y)=(x^2+y^2)e^{-(x^2+y^2)}$$ attains a maximum at every point of the unit circle. The determinant of the hessian matrix at those points is zero so I wrote $f$ in polar coordinates and tryed to prove there is no $r \in \mathbb{R}$ satisfying $$r^2e^{-r^2}>e^{-1}$$ which is equivalent to $$e^{r^2}<er^2$$ I don't know how to prove it, any suggestion is welcome.

Thanks

Your statement is false for $0\leq r<1$ (and you don't need to consider $r<0$ since the domain is restricted due to the coordinate transformation). — LinAlg, Jan 18 '17 at 18:32
Recall: $e^z\geq 1+z$ for $z\in\mathbb{R}$. Use this with $z=r^2-1$. — yurnero, Jan 18 '17 at 18:49

score 1 · Answer 1 · answered Jan 18 '17 at 18:42

1

The function $g(r) = e^r$ is strictly convex, therefore its graph lies above each tangent line: $$ e^r = g(r) \ge g(1) + g'(1)(r-1) = er $$ with equality only for $r=1$.

answered Jan 18 '17 at 18:42

Martin R

113,040

score 0 · Accepted Answer · answered Jan 18 '17 at 18:30

0

This can be proven more easily. Consider $$f(z) = z e^{-z} (z\geq 0) \quad f'(z) = (1-z)e^{-z}$$ The derivative is $0$ only for $z=1$. Since $f'(z) > 0 $ for $z<1$ and $f'(z) < 0$ for $z>1$, $z=1$ corresponds to a unique maximum.

answered Jan 18 '17 at 18:30

LinAlg

19,822

Of course, didn't realize I had to think it as a one variable extrema problem. Thanks! – la flaca Jan 18 '17 at 18:39

How to prove that $er^2 \leq e^{r^2}$ for all $r \in \mathbb{R}$

2 Answers2