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I know that the geometric interpretation of differentiability for a function $f:\mathbb{R}^2\to \mathbb{R}$ in a point $(x_0,y_0)$ is that it admits a tangent plane in the point $P=(x_0,y_0,f(x_0,y_0))$. Thinking about the concept geometrically, I came up with the following ""conjecture"". Let for simplicity $\ell_{v}$ the tangent line in the point $P$ to the graphic of $f$ in the direction of the versor $v$, and let's suppose the function $f$ admits directional derivatives in every direction.

$f \text{ differentiable in } (x_0,y_0)$ $\iff$ $\forall v,u \ \ \text{versors}$, $\ell_{v}$ $\text{and}$ $\ell_{u}\ \text{are coplanar}$

I tried proving this using affine geometry but calculations get pretty messy. I would like if it's true or not(in the former case I would like a proof, in the latter a counterexample)

Rodrigo de Azevedo
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Kandinskij
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1 Answers1

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This is borderline duplicate of Continuous function with linear directional derivatives=>Total differentiability?, but restated in geometric language. The fact that $l_v$ exist is equivalent to directional derivatives $f_v$ existing. The fact that they are coplanar is equivalent to the function $L(v)=f_v$ being linear. So the counterexamples in that question are counterexamples for your question as well.

Geometrically, the tangent plane has to approximate the graph "uniformly well" over a disc in $\mathbb{R}^2$, while tangent line each approximates the graph over a segment; when, for a "given quality of approximation", these segments fail to assemble into a disc (as in the counterexample given), the resulting plane will fail to approximate, and will not be the tangent plane in the usual sense (of total differentiability).

Max
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