Proof or counter-example that $(0,t_0)$ is a maximum of $f: [0,1]^2 \to \mathbb{R}$ if $t_0$ is a maximum of $g(t) := f(0,t)$.

Question

Suppose $f: [0,1] \times [0,1] \to \mathbb{R}$ is a differentiable function. Define $g(t) = f(0,t)$ and suppose that $g$ has a maximum in $t_0 \in (0,1)$, and suppose additionally that $D_1f(0,t_0) < 0$. I need to either prove, or give a counter-example to the fact that then $(0,t_0)$ is a local maximum of $f$.

Firstly it follows that $g'(t_0) = D_2f(0,t_0) = 0$. I have set up the Taylor expansion, write $\vec{a} = (0,t_0)$, then: $$ f(\vec{a} + \vec{h}) = f(\vec{a}) + D_1f(\vec{a})h_1 + D_2(\vec{a})h_2 + o(\|\vec{h}\|) = f(\vec{a}) + D_1f(\vec{a})h_1 + o(\|\vec{h}\|). $$ Now if $h_1 < 0$ it seems that $f(\vec{a} + \vec{h}) > f(\vec{a})$, or at least has the potential to be greater, so my guess is that it does not hold and we need to specify a counter-example. I have tried several but everything fails.

Answer 1

Let's build $f$ such that $g(t) = t-t^2$ (which attains its maximum at $t_0 = \dfrac12$) but which increases with $x$.

For this purpose, let's try $f(x,y) = x + y - y^2$. When we fix a value for $x$, the graph is simply the same parabola but translated to the top by $x$ (see below the drawing in $\Bbb R^3$ of the surface $z = f(x,y)$).

In this case, $f(0,0.5) = g(0.5) = 0.25$ but $f(1,0.5) = 1.25$ so that $(0,0.5)$ isn't a point of maximum of $f$.

Edit: I did not see the condition $D_1f(0,t_0)<0$. Thus I propose a new counterexample $f(x,y) = x^2-\dfrac12 x+y-y^2$ (see the new picture below).

Here, the idea was to add a function of $x$ (that is, $x^2-x$ here) which begins by decreasing when $x$ increases from 0 (in particular, $D_1f(0,0.5) = -\dfrac 12<0$) and then has a sufficiently strong increase when $x$ approaches $1$ so that $f(1,0.5)>f(0,0.5)$.

Answer 2

This example is built around a standard example of a function for which both partials exist at a point, yet the function behaves badly near the point. That will be called $q(x,y),$ and is $(xy)/(x^2+y^2)$ except at the origin where it is defined as $0.$ One can then show both partials of $q$ are $0$ at the origin.

For ease of discussion we redefine the domain to be $[-1,1] \times [0,1]$ and will take $t_0=0$ noting that this is in the open interval $(-1,1).$

Now define $f(x,y)=q(x,y)-x-y^2.$ We have $D_1(0,0)=-1<0$ as required. Also $g(t)=f(0,t)=-t^2$ which has its maximum at $0 \in (-1,1)$ where it is $0.$

Now $f$ fails to have a local maximum at the origin, since (for $t \neq 0$) $f(t,t)=1/2-t-t^2,$ which tends to $+1/2$ as $t \to 0.$

Edit: This example is not differentiable at the origin (the interior point of the left side of the altered domain).