Extrema of $f(x):=e^{x(y+1)}$ on $\mathcal C_1^{\|\cdot\|_\infty}$

Question

$$f(x):=e^{x(y+1)}$$

Are the extrema on $\mathcal C_1^{\|\cdot\|_\infty}$ where $\mathcal C_1^{\|\cdot\|_\infty}$ denotes the closed circle around $(0,0)$ with radius $1$ (sup norm!) the same as on $\mathcal C_1^{\|\cdot\|_2}$?

$\mathcal C_1^{\|\cdot\|_2} =\{(x,y) \in \mathbb{R}^2:\sqrt{x^2+y^2} \le 1\}$

Answer 1

Hint

The unit circle defined with the sup norm is more of a square than a circle. $\hskip2in$

Focus your attention on what the function is in the upper right plane $x\gt 0$ $y\gt 0$ and in the lower left plane $x\lt 0$ $y\lt 0$

Answer 2

If you are asking in general, then the answer is no, the extrema of some function $f$ on different balls are indeed not the same.

If you are asking particularly about $C_1^{\| \cdot \|_2}$ and $C_1^{\| \cdot \|_{\infty}}$ then the answer is again no. These norm have almost nothing in common, some results on much stronger assumptions may imply for dual norms, say $\| \cdot \|_1$ and $\| \cdot \|_{\infty}$.

If you are asking for the function $f(x, y) = e^{x(y + 1)}$ then I'm afraid the answer is again no. It is fairly easy (tell me if you have difficulties) to solve the following optimization problem using Lagrange multiplier method $$ \text{ optimize } f(x, y) := e^{x(y + 1)} \\ \text{ subject to } x^2 + y^2 \le 1. $$ It's even easier to find the extrema points of above problem, which are $(0, -1), \left(\frac{\sqrt{3}}2, \frac 12\right), \left(-\frac{\sqrt{3}}2, \frac 12\right), \left(\frac{\sqrt{3}}2, -\frac 12\right), \left(-\frac{\sqrt{3}}2, -\frac 12\right)$. The global maximum is achieved at $\left(\frac{\sqrt{3}}2, \frac 12\right)$ with function value $e^{\frac{3\sqrt{3}}{4}}$ and the global minimum is achieved at $\left(-\frac{\sqrt{3}}2, \frac 12\right)$ with function value $e^{-\frac{3\sqrt{3}}{4}}$.

Now consider the problem with sup-norm:

$$ \text{ optimize } f(x, y) := e^{x(y + 1)} \\ \text{ subject to } \max(x, y) \le 1. $$ The problem is still convex and can be solved analytically, though using a bit different technique. Since, $\log(\cdot)$ is a monotonic function then its equivalent to taking the logarithm of objective function and the maximum/minimum points will not be changed. Namely, $$ \text{ optimize } g(x, y) := {x(y + 1)} \\ \text{ subject to } \max(x, y) \le 1. $$ has the same set of maximum/minimum points. Plotting the square $\max(x ,y) = 1$ and the hyperbola $x (y + 1) =c \implies y = \frac cx - 1$ we see that the maximum possible value for $c$ is $2$ when the hyperbola and square intersect in the upper right corner of the square, namely $x = y = 1$. The maximum point hence is $(1, 1)$ with maximum value of $e^2$, which is indeed different from $e^{3\sqrt{3}/4}$ from the previous setup.