Let's try this again.
Let $f:D\to\mathbb{R}$, $D\subset\mathbb{R}^2$. Let $X\in D$. By a derivative in the direction of nonzero vector $\bf{v}$ we mean $$\lim_{h\to 0}\frac{f(X+h{\bf{v}})-f(X)}{h} $$ if such a limit exists we'll denote the result as $\mbox{D}_{\bf{v}}f(X)$
Note that when we pick direction as vector $(1,0)$ and $(0,1)$ we obtain the partial derivatives at point $X$. It is also known that a directional derivative can be computed by a certain scalar product: $$\mbox{D}_{\bf{v}}f(X) = \left\langle \nabla f(X), \frac{\bf{v}}{\|\bf{v}\|} \right\rangle $$

The problem
Let $D\subset\mathbb{R}^2$, $f:D\to\mathbb{R}$ be continuous at $A\in\mbox{Int}D$ and let there exist $r>0$ s.t $$X\in B(A,r), X\neq A \Longrightarrow \mbox{D}_{\bf{AX}}f(X) >0 \tag{*}\label{eq:1}$$ Can we assume that this implication guarantees the existence of derivative in any direction?

Show that for any $X\in B(A,r), X\neq A$ we have $f(X)>f(A) $.

In a single variable case, if we have $g:[a,b]\to\mathbb{R}$ with $g$ continuous at $a,b$ and $g'(x)>0$ for any $x\in (a,b)$ we can immediately say that $g$ is differentiable in $(a,b)$ and therefore for any $x\in (a,b]$ we can find $c\in (a,x)$ such that $$g(x)-g(a) = g'(c)(x-a)>0 $$ and conclude that $g(a)$ is the minimum.

In multiple variable case, my hope is to achieve something similar. To be precise, the mean value theorem is as follows:
Let $g:E\to\mathbb{R}$, $E\subset\mathbb{R}^2$ be continuous at $X, X+v$, $v:=(v_1,v_2)$ and differentiable in the straight line $[X, X+v]$ that connects these points. Then there exists $C\in (X,X+v)$ such that $$g(X+v)-g(X) = g_x(C)v_1+g_y(C)v_2 $$ The proof doesn't actually utilize the dimension of the domain, so I suppose it remains valid for any arbitrary finite dimensional domain

An attempt
The following assumes that \eqref{eq:1} guarantees existence of derivatives in arbitrary direction
By \eqref{eq:1}, for any $X\in B(A,r), X\neq A$ we have $\mbox{D}_{\bf{AX}}f(X)>0$, which means $$\left\langle \nabla f(X),\frac{\bf{AX}}{\|\bf{AX}\|}\right\rangle >0\Longrightarrow f_x(X)(x-a)+f_y(X)(y-b)>0 $$ where $X = (x,y), A = (a,b)$.
If $f$ was differentiable in the entire $B(A,r)$, the the result would be immediate from MVT. For diffentiability, it is sufficient for $f$ to have continuous partial derivatives, but we only know they exist at every point, not that they are necessarely continuous (right?).

An idea
If we pick an arbitrary direction w.r.t to point $A$, let's say it's determined by the vector ${\bf{u}}\neq 0$, then we could view a "cut" or a "section" of the function $f$ which we could define by $$f_{\bf{u}}(t) := f(a+tu_x, b+tu_y) $$ imagine we slice the surface $f$ with a plane that is perpendicular to the $x,y$ plane, passes through $A$ and follows the direction of $\bf{u}$
However, we must choose the parameter $t$ such that $A+t\bf{u}$ leaves point $A$ in only one direction.
The derivative of such a function at some point $A+t\bf{u}$ is precisely the directional derivative of $f$ at $A+t{\bf{u}}$ and now we can apply single variable MVT for the entire family of such functions and conclude that $f(A)$ has to be the minimum.


Hint:This follows from the $1$-dimensional case, consider $h_X:I\rightarrow R^n$ defined $h_X(t)=A+tX$ where $h_X(I)$ is in the domain of $f$ and $f(h_X(t))$, its derivative at $0$ $d_{AX}f>0$. This implies that $h\circ f$ increases in $I$, remark that $A$ is not a minimum, because you cannot have $d_{AX}f(X)>0$ and $d_{A(-X)}f=-d_{AX}f>0$. In fact you can characterize local extremums like in the $1$-dimensional case here if $f$ is defined on $B(A,r)$, and is differentiable, then $df_{AX}f=0$ for every $X$ if $A$ is a local minimum.

For the situation $g:[a,b]\rightarrow R$ with $g'(x)>0, x\in (a,b)$ $a$ can be a minimum because $a$ is not in the interior of the domain of $g$.

  • $\begingroup$ Oh, that is clever! $\endgroup$
    – AlvinL
    Mar 4 '17 at 10:35

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