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A famous theorem says that gradient is perpendicular to the level set. Indeed, wiki says

If the function $f$ is differentiable, the gradient of $f$ at a point is either zero, or perpendicular to the level set of $f$ at that point.

The problem is that in every textbook I have the formulation of this theorem (and hence, its proof) requires function $f$ to be differentiable in every point of its domain.

  1. Can we relax this requirement and reformulate this theorem as follows:
    If function $f: X \subset \mathbb{R}^n \to \mathbb{R}$ is differentiable at point $\mathbf{x}_0 \in \operatorname{int} X$ and $\nabla f(\mathbf{x}_0) \neq \mathbf{0}$, then its gradient $\nabla f(\mathbf{x}_0)$ is perpendicular to the level set $L(f(\mathbf{x}_0))$ at point $\mathbf{x}_0$?

  2. If the answer to the question above is "yes" then can we in proof of this theorem (you can find it here, for example) relax the requirement "let $\gamma: (-a,a) \to \mathbb{R}^n$ be a $C^1$ curve contained in $L(f(\mathbf{x}_0))$ and such that $\gamma(0) = \mathbf{x}_0$"
    to the following local requirement "let $\gamma: (-a,a) \to \mathbb{R}^n$ be a curve that is continuously differentiable at point $0$ and contained in $L(f(\mathbf{x}_0))$ and such that $\gamma(0) = \mathbf{x}_0$" ?

Rodvi
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1 Answers1

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tl; dr: The answer to 1. is emphatically no.

Let $\phi$ is an arbitrary bounded, real-valued function is a neighborhood of the origin (in the plane, say), and define $$ f(x, y) = x + (x^{2} + y^{2})\phi(x, y). $$ It's straightforward to check $f$ is differentiable at $(0, 0)$ and $\nabla f(0, 0) = (1, 0)$.

On the other hand, the zero set of $f$ can be arbitrary by taking $\phi$ a characteristic function.

Andrew D. Hwang
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