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I want to show that if a map $f$, defined on an open subset $X$ of a Banach space $E$ to another Banach space $Y$, is differentiable at a point $x_0 \in X$ then the directional derivative with respect to a nonzero vector $v \in E$ of $f$ at $x_0$ also exists. For this purpose, I will not assume the definition of the directional derivative but infer it from the consequences of the hypotheses above. There are two basic questions about this I would like to ask: Is my argument sound and is there a more concise argument that lead to the same conclusions?

Now, assuming the above hypotheses, since $X$ is open we can take a sufficiently small $t$ such that $x_0 + tv$ lies in $X$. That is, there exists $\epsilon >0$ such that $x_0 + tv \in X$ for $|t| < \epsilon$. This means that the function $$ g(t) = f(x_0 + tv) $$ is defined on a neighborhood of $x_0$

Denote the derivative of $f$ at $x_0$ by $\partial f(x_0)$. Then, there exists a continuous map $r:X \rightarrow F$ such that $r(x_0) = 0$ and

$$ f(x) = f(x_0) + \partial f(x_0)(x - x_0) + r(x)||x - x_0|| $$

Then,

$$ g(t) = f(x_0 + tv) $$

$$ = f(x_0) + \partial f(x_0)( (x_0 + tv) - x_0) + r(x_0 + tv)||(x + tv) - x_0|| $$

$$ = f(x_0) +\partial f(x_0)(tv) + r(x_0 +tv)||tv|| $$

By homogeneity of the norm and linearity of the derivative, it follows that

$$ f(x_0 + tv) = f(x_0) + t \partial f(x_0)(v) + |t|r(x_0 + tv)||v|| $$

From algebraic manipluation, it follows that

$$ \frac{f(x_0 + tv) - f(x_0)}{t} - \partial f(x_0)(v) = \frac{|t|}{t} ||v||r(x_0 + tv) $$

Note that as $t \rightarrow 0$ the right hand side of the equation approaches $0$ since $\frac{|t|}{t}||v|| = ±||v||$ is constant and

$$ \lim_{t \to 0} \; r(x_0 + tv) = r(x_0) = 0 $$

where the first equality follows from the continuity of $r$ and the last follows from differentiability of $f$ as noted above.

Therefore,

$$ \lim_{t \to 0} \frac{f(x_0 + tv) - f(x_0)}{t} = \partial f(x_0)(v) = D_v f(x_0) $$

where we obtain in this last expression the usual definition of the directional derivative $D_v f(x_0)$ of a function $f$ at $x_0$ with respect to the vector $v$.

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1 Answer 1

up vote 2 down vote accepted

The argument is correct overall, but you are essentially re-proving a special case of the Chain Rule. If you're willing to assume the Chain Rule, the following argument is a bit shorter:

Let $\gamma\colon (-\epsilon,\epsilon)\to X$ be the function defined by $$ \gamma(t) \;=\; x_0 + vt\text{,} $$ and note that $g = f\circ \gamma$. Clearly $\gamma$ is differentiable, with $\gamma'(0) = v$. By the Chain Rule, it follows that $g$ is differentiable at $0$, with $$ g'(0) \;=\; (f\circ \gamma)'(0) \;=\; \partial f(x_0)(\gamma'(0)) \;=\; \partial f(x_0)(v). $$

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I agree that this argument is shorter - just count the lines! - but you use the chain rule, which is (as you say) established by essentially the computation made by the OP. –  t.b. May 28 '11 at 19:11

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