4
$\begingroup$

Let $X$ and $Y$ be normed real vector spaces, and $f : X \to Y$ a map.

Let's say that:

G) $f$ is Gateaux differentiable at $x_0 \in X$ if for all directions $v \in X$ the limit $f'(x_0)(v) := \lim_{t \searrow 0} t^{-1} [f(x_0 + t v) - f(x_0)]$ exists in $Y$.

F) $f$ is Frechet differentiable at $x_0 \in X$ if there exists a bounded linear operator $A : X \to Y$ such that for all $h \in X$: $f(x+h) - f(x) = A h + o(\|h\|_{X})$ as $h \to 0$ in $X$. That's a little Landau-oh; the residual term $o(\|h\|_{X})$ is supposed to be bounded in terms of $\|h\|_X$ only, uniformly in $h / \|h\|_X$.

Fix $x_0 \in X$.

Once we know that $f$ is Gateaux differentiable at $x_0$, there are several obstructions to it being Frechet differentiable at $x_0$.

Suppose $f$ is Gateaux differentiable at $x_0$ and the Gateaux differential $v \mapsto f'(x_0)(v)$ is one of the following:

0) exactly the Frechet derivative.

1) linear but not continuous in $v$.

2) continuous but not linear in $v$.

3) linear and continuous in $v$.

Suppose the Gateaux derivative $f'(x) : X \to Y$ is a bounded linear functional for each $x \in X$ and $f'$ is one of the following:

4) continuous at $x_0$: $f'(x) \to f'(x_0)$ in operator norm whenever $x \to x_0$ in $X$.

5) continuous at $x_0$ weakly: $f'(x) \to f'(x_0)$ in operator norm whenever $x$ converges weakly to $x_0$.

6) weakly continuous at $x_0$: for each $v \in X$ fixed, one has $f'(x)(v) \to f'(x_0)(v)$ whenever $x \to x_0$ in $X$.

You are welcome to answer any subset of the following questions.

a) Is any of those notions redundant?

b) Is any of those notions redundant if $X$ and $Y$ are Banach spaces?

Provide examples, if such exist, for:

c): 3) but not 0).

d): 6) but not 5).

e): 4) but not 5).

f): $x \mapsto f'(x)$ is $C^1$ but not 0).

g): where 5) occurs.

h): where 6) occurs.

Suppose in 4)--6), $f$ is Frechet differentiable at each $x$.

i) In which case is $f$ necessarily Frechet differentiable at $x_0$?

j) Same as i), but assuming $X$ and $Y$ are Banach spaces.

$\endgroup$
1
$\begingroup$

Here are some answers

c): 3) but not 0) - This is not possible. A continuous linear operator is it's own Frechet derivative. Where the Frechet derivative exists it coincides with the Gateaux derivative.

f): 4) but not 5) - This is not possible. When the derivative in the normal sense exists then the Frechet and Gateaux derivatives exist and coincide

If you can get access to the book 'Nonlinearity and Functional Analysis, Berger, 1977' there is a nice introduction to Frechet and Gateaux differentiability. A statement that will be useful for some of the questions you have asked is given in section 2.1:

'If the Gateaux derivative $df(x_{0},h)$ is linear in $h$ (i.e. $df(x_{0},\cdot) \in L(X,Y)$ and is continuous in $x$ as a map from $X \rightarrow L(X,Y)$ then $f$ is Frechet differentiable at $x_{0}$'. The Frechet and Gateaux derivative coincide in this case

Sorry I haven't answered any of the other questions, but I hope this helps

$\endgroup$
  • $\begingroup$ I don't think, c) is addressed adequately. I didn't understand the second subanswer; was it about f) or e)? In any case, I don't know what is the "normal sense" you are referring to. $\endgroup$ – user66081 Feb 4 '14 at 14:45
1
$\begingroup$

I will collect answers in this post.

c): 3) but not 0).

Take $X$ as the space of finitely supported sequences with the $|\cdot|_\infty$ norm, and $f : X \to \mathbb{R}$, $x \mapsto \sum_{n \in \mathbb{N}} n x_n^2$. Then $f'(0) = 0$ is the Gateaux differential. But $f(0+h) - f(0) \geq 1$ for $h = r e_n$, where $n r^2 \geq 1$, and $|h|_\infty = r > 0$ may be arbitrary small. This defies the existence of the Frechet derivative at $0$.

f): $x \mapsto f'(x)$ is $C^1$ but not 0).

From the reference pointed out by Keeran Brabazon (Berger, Nonlinearity and functional analysis, 1977): If the Gateaux differential $f'$ is continuous at $x_0$ as a mapping $X \to L(X; Y)$, the latter being the space of bounded linear operators $X \to Y$, then $f$ is Frechet differentiable at $x_0$. So f) is not possible.

i)+4): see f).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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