Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


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

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

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

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.


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

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 $v$.

share|cite|improve this question
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 $$ g'(0) \;=\; (f\circ \gamma)'(0) \;=\; \partial f(x_0)(\gamma'(0)) \;=\; \partial f(x_0)(v). $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.