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Let $f:X \to Y$ be a (nonlinear) function between Hilbert spaces that such that $$f(x+th)-f(x) = tD + r(t)$$ for small $t>0$, holds for a function $r$ which satisfies $t^{-1}r(t) \to 0$ as $t \to 0$.

Suppose that $b$ is such that $t^{-1}b(t) \to 0$ as $t \to 0$. Is it possible to derive an expression for $$f(x+th+b(t))$$ in terms of the above expression involving $D$ and $r$? I guess we may need $f$ to be differentiable in all directions, which is fine. I guess I am asking whether if the directional derivative of $f$ exists at $x$, then does also the limit $$\lim_{t \to 0}\frac{f(x+th+b(t))-f(x)}{t}$$ exist?

Since $b(t)$ behaves like $const\times t$, I thought maybe this is true.

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    I have a question about multiplying element of different Hilbert spaces. I mean what does the following mean $t^{-1}r(t)$, or $tD$ – 2017-01-11
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    @kolobokish Check definition of a vector space – 2017-01-11
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    @StopUsingFacebook What you mean, by that? I know the definition of vector space. I don't know what is understood under the multiplication of two element of some different two vector spaces. That has nothing to do with the definition of one vector space. It should be spacified if $t$ is taken from underlying field. And that $r(t)$ is a map from $R$ to $Y$. – 2017-01-11

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If $f$ is Fréchet differentiable at $x$, then of course we have

$$f(x + th + b(t)) = f(x) + Df\lvert_x(th + b(t)) + r(t) = f(x) + t\cdot Df\lvert_x(h) + o(t)$$

whenever $b\colon \mathbb{R} \to X$ satisfies $\lim\limits_{t\to 0} t^{-1}b(t) = 0$.

If $f$ is only assumed Gâteaux differentiable at $x$, even if the Gâteaux differential is a continuous linear operator, bad things can happen, and

$$\lim_{t\to 0} \frac{f(x + th + b(t)) - f(x)}{t}$$

need not exist.

For an example, consider $X = \mathbb{R}^2$ and $Y = \mathbb{R}$. Let $g \colon [0, +\infty) \to \mathbb{R}$ be given by

$$g(x) = \max \:\bigl\{0, 1 - \lvert x-2\rvert\bigr\},$$

and

$$f(r\cos \varphi, r\sin \varphi) = g\bigl(r\tan\tfrac{\varphi}{4}\bigr),$$

where $r \geqslant 0$ and $\varphi \in [0,2\pi)$.

Then for every $\varphi \in [0,2\pi)$ there is a $\varepsilon > 0$ such that $f(t\cos \varphi, t\sin \varphi) = 0$ for all $t\in (-\varepsilon, +\varepsilon)$, so $f$ is Gâteaux differentiable at $0$ with directional derivative $0$ in all directions. But $f$ isn't even continuous at $0$, and if we take $h = (1,0)$ and $b(t) = t\tan (2t)\cdot (0,-1)$, then

$$\lim_{t\to 0^+} f(th + b(t)) = 1,$$

so the difference quotient blows up.

Using a smooth bump function instead of the piecewise linear $g$ from above, and a suitable amplitude depending on the angle, we can construct an example exhibiting essentially the same behaviour that is smooth ($C^{\infty}$) on $\mathbb{R}^2\setminus \{0\}$, so requiring global Gâteaux differentiability wouldn't help. For good local behaviour, one needs Fréchet differentiability.

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    Thanks. If $f$ is Lipschitz, it works, no? One can add and subtract $f(x+th)$ and show that the problem term vanishes in the limit. – 2017-01-12
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    Yes, if $f$ is Lipschitz, it works. It also works if we assume Hadamard directional differentiability (I just looked up what that is), since stability against small perturbations of the direction are built into that. – 2017-01-12
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    @DanielFischer If only Hadamard, your first displayed equation doesn't necessarily hold does it? Only if the derivative is linear. – 2017-02-14
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    @25Chars We have equality of the outer terms if $f$ is Hadamard-differentiable in direction $h$ at $x$, provided we interpret $Df\lvert_x(h)$ as the Hadamard derivative, and restrict $t$ to positive values. If the Hadamard derivative is linear, the middle term also works, but otherwise … not so much. – 2017-02-14
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    @DanielFischer Do you mean that the LHS equals the RHS (but not equal to the middle hand side) of your displayed equality? I didn't know what you mean by outer terms and middle term. I would've thought that the LHS = middle, and this equals the RHS if eg. we have linearity of the derivative. – 2017-02-14
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    @25Chars If $f$ is Hadamard differentiable at $x$ in direction $h$, and $b(t) \in o(t)$, then almost by definition of the Hadamard derivative we have (for $t > 0$) $$f(x + th + b(t)) = f(x) + t\cdot D^Hf\lvert_x(h) + o(t).$$ The Hadamard derivative is (at least on the [wikipedia page](https://en.wikipedia.org/wiki/Hadamard_derivative)) defined in terms of sequences, so we must still make the step from sequences to a continuous parameter to prove that representation, but that's a standard argument. – 2017-02-14
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    For the middle term to make sense with Hadamard derivatives, $f$ must be Hadamard differentiable in direction $th + b(t)$ at $x$. Since we have homogeneity, that is equivalent to Hadamard differentiability in direction $h + t^{-1}b(t)$, and we can write the middle term as $f(x) + t\cdot D^Hf\lvert_x(h + t^{-1}b(t)) + r(t)$ if $D^Hf\lvert_x(h+t^{-1}b(t))$ exists. But even if all these Hadamard derivatives exist, it's at least not clear how well-behaved the Hadamard derivative is under small changes of direction. – 2017-02-14