Prob. 22, Exercises 8.9, in Apostol's CALCULUS vol 2: There is no scalar field with all non-zero directional derivatives at a single point

Question

Here is Prob. 22, Exercises 8.9, in the book Calculus Volume II by Tom M. Apostol, 2nd edition:

(a) Prove that there is no scalar field $f$ such that $f^\prime(\mathbf{a}; \mathbf{y})>0$ for a fixed vector $\mathbf{a}$ and every nonzero vector $\mathbf{y}$.

(b) Give an example of a scalar field $f$ such that $f^\prime(\mathbf{x}; \mathbf{y}) > 0$ for a fixed vector $\mathbf{y}$ and every vector $\mathbf{x}$.

The example for part (b) is the scalar field $f$ defined on $\mathbb{R}^n$ by the formula $$ f(\mathbf{x}) = \langle \mathbf{x}, \mathbf{y} \rangle \ \mbox{ or } \ \mathbf{x} \cdot \mathbf{y} \ \mbox{ for all } \ \mathbf{x} \in \mathbb{R}^n. $$ For in this case, we note that $$ \frac{f(\mathbf{x} + h \mathbf{y} ) - f(\mathbf{x}) }{ h} = \mathbf{y} \cdot \mathbf{y} = \lVert \mathbf{y} \rVert^2, $$ and so $$ f^\prime( \mathbf{x}; \mathbf{y}) = \lVert \mathbf{y} \rVert^2 > 0, $$ for every nonzero vector $\mathbf{y}$.

How to tackle Part (a)?

Here is Apostol's definition of the directional derivative:

Given a scalar field $f \colon S \to \mathbb{R}$, where $S \subseteq \mathbb{R}^n$. Let $\mathbf{a}$ be an interior point of $S$ and let $\mathbf{y}$ be an arbitrary point in $\mathbb{R}^n$. The derivative of $f$ at $\mathbf{a}$ with respect to $\mathbf{y}$ is denoted by the symbol $f^\prime(\mathbf{a}; \mathbf{y})$ and is defined by the equation $$ f^\prime(\mathbf{a}; \mathbf{y}) = \lim_{h \to 0} \frac{ f( \mathbf{a} + h \mathbf{y} ) - f( \mathbf{a})}{h} $$ when the limit on the right exists.

By this definition, we can conclude that if one of $f^\prime(\mathbf{a}; \mathbf{y}) $ and $f^\prime(\mathbf{a}; - \mathbf{y}) $ exists, then the other exists also, and the two are equal. Am I right?

If so, then how to proceed in proving the assertion in Part (a)?

Answer 1

Albeit, in the definition there isn't restriction about $h$, almost ever we suposse it positive. We travel from $\mathbf{a}$ to $\mathbf{a + y}$ albeit the path is shorter and shorter when $h\to 0$.

You can, like sugest saulpatz, replace $\mathbf{y}$ by $-\mathbf{y}$ (by hypothesis about $\mathbf{y}$), next make a substitution $l=-h$, this do look the definition like the original, and then you arrive to $0<f'(\cdot)=-f'(\cdot)<0$, a contradiction.

Answer 2

I think you've dropped a minus sign somewhere. $$ f^\prime(\mathbf{a}; \mathbf{-y}) = \lim_{h \to 0} \frac{ f( \mathbf{a} - h \mathbf{y} ) - f( \mathbf{a})}{h} = \lim_{h \to 0} \frac{ f( \mathbf{a} + h \mathbf{y} ) - f( \mathbf{a})}{-h}=-f^\prime(\mathbf{a}; \mathbf{y}) $$