I am new to differential calculus on normed spaces and I struggle with some easy things.

Let $- \infty <a < b< +\infty$ and $[a, b]$ denote a finite interval.

Let $C[a,b]$ denote the collection of all real-valued continuous functions defined on $[a,b]$. Then, endowed with the usual choice of norm $\|x\| = \max_{a\leq t \leq b} |x(t)|$, $C[a,b]$ is a Banach space.

Let $\phi \colon \mathbb{R} \to \mathbb{R}$ be a twice differentiable function, and suppose that its inverse $\phi^{-1} \colon \mathbb{R} \to \mathbb{R}$ exists and is still twice differentiable.

Let the kernel function $K \colon [a,b] \times [a,b] \to \mathbb{R}$ be continuous.

Does the Frechet derivative of the following operator $A$ exist? $$ [A(x)](s) = \int^b_a K(s,t) \, [ x(t)] \,\mathrm{d}t $$ for all $x \in C[a,b]$ and for all $s \in [a,b]$.

Any ideas or suggestions are much appreciated! Thanks in advance:)

  • $\begingroup$ Do you mean $$[A(x)](s) = \phi^{-1} \left( \int^b_a K(s,t) \,\phi [ x(t)] \,\mathrm{d}t \right)$$ without $\circ$? $\endgroup$ – md2perpe Jan 12 '18 at 22:28

The other answer is correct, but shows only that the map is Gateaux differentiable as you have noted.

To see that it is Frechet differentiable (this is commonly just called "differentiable"), we write the map as a composition:

$$A=L^{-1}\circ \hat K\circ L,$$ here $L(f)=\phi\circ f$ (so $L^{-1}(f)=\phi^{-1}\circ f$) and $\hat K(f)\,(t)=\int_a^b K(t,y)f(y)\,dy$. The map $\hat K$ is linear and continuous, thus differentiable with differential $(d_f \hat K)\,(g)=\hat K(g)$.

We will verify differentiability of the map $L$, the map $L^{-1}$ is basically the same map, except with $\phi$ replaced by $\phi^{-1}$. Your conditions on $\phi$ and $\phi^{-1}$ are the same (both must be twice differentiable functions $\Bbb R\to \Bbb R$), so a proof of the differentiability of $L$ is also a proof of the differentiability of $L^{-1}$.

Once we have found all the differentials the chain rule will tell us:

$$d_fA=d_{\hat K(L(f))}(L^{-1})\cdot d_{L(f)}\hat K\cdot d_f L$$

So why is $L$ differentiable? Let $M_{\phi'\circ f}$ be the map $M_{\phi'\circ f}(g)\,(t)=\phi'(f(t))\cdot g(t)$. This is, for any $f\in C[a,b]$, a continuous linear map $C[a,b]\to C[a,b]$.

Now lets use the mean value theorem, which tells us that $$\phi(f(t)+h(t))-\phi(f(t)) = \phi'(\xi)\cdot h(t)$$ for some $\xi\in [f(t),f(t)+h(t)]$. Now $\phi'$ is itself continuously differentiable, so $$|\phi'(\xi)-\phi'(f(t))|≤\sup_{x\in[a,b]}|\phi''(f(x))|\cdot |f(t)-\xi|≤\|\phi''\circ f\|\,\|h\|.$$ It follows: \begin{align} \frac{\|L(f+h)-L(f)-M_{\phi'\circ f}(h)\|}{\|h\|} &=\frac{\sup_{t\in[a,b]}|\phi(f(t)+h(t))-\phi(f(t))-\phi'(f(t))\cdot h(t)|}{\|h\|}\\ &=\sup_{t\in[a,b]}\frac{|\phi'(\xi)-\phi'(f(t))|\,|h(t)|}{\|h\|}≤\|\phi''\circ f\|\,\|h\|. \end{align} So as $\|h\|\to0$ this expression goes to $0$. This is precisely the condition that $L$ is differentiable at $f$ with differential $d_f L=M_{\phi'\circ f}$. The differential of $L^{-1}$ at $f$ is $d_f(L^{-1})= M_{(\phi^{-1})'\circ f}=M_{\frac1{\phi'}\circ f}$.


Let $\hat K : C[a,b] \to C[a,b]$ be the linear map defined by $$\hat K(f) = \int_a^b K(\cdot, t) \, f(t) \, dt.$$

Then we have $A(x) = \phi^{-1} \circ \hat K(\phi \circ x)$ and $$ A(x+\lambda h) = \phi^{-1} \circ \hat K(\phi \circ (x+\lambda h)) $$

Expanding in $\lambda$ we now have $ \phi \circ (x + \lambda \, h) \sim (\phi \circ x + \lambda \, h \, \phi' \circ x) $ so $$ \hat K(\phi \circ (x+\lambda h)) \sim \hat K(\phi \circ x + \lambda \, h \, \phi' \circ x) = \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x) $$ by linearity of $\hat K.$

Then $$ A(x+\lambda h) \sim \phi^{-1} \circ \left( \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x) \right) \\ \sim \phi^{-1} \circ \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x) \cdot (\phi^{-1})' \circ \hat K(\phi \circ x) \\ $$

Thus, $$ \left< DA(x), h \right> = \hat K(h \, \phi' \circ x) \cdot (\phi^{-1})' \circ \hat K(\phi \circ x). $$

  • $\begingroup$ Thank you so much @md2perpe. May I confirm one little thing with you please? I think your answer shows the result of the Gateaux derivative explicitly. But our purpose is to figure out the Frechet derivative. In fact, as we know, there is a slight difference between Gateaux and Frechet derivative, unless the G-derivative at $x$ for which the passage to the limit (i.e., $\left<DA(x), h \right> = \lim_{t \to 0} \frac{A(x + \lambda h) - A(x)}{t}$) is uniform for all $h$ with $\|h\|=1$. Thus, would you mind to explain more about that please? Many thanks again:) $\endgroup$ – Paradiesvogel Jan 13 '18 at 10:59
  • $\begingroup$ According to a paragraph on Wikipedia the Gâteaux derivative of functionals between complex Banach spaces is automatically linear and equals the Frechet derivative (which then also exists). In your case you have limited the spaces to real values, but it's obvious that the case can easily be extended to complex values without changing the result. $\endgroup$ – md2perpe Jan 13 '18 at 14:22
  • $\begingroup$ Thanks @md2perpe. Sorry, I'm still confused. There is a quote from Wikipedia you cited, "If F is Fréchet differentiable, then it is also Gâteaux differentiable, and its Fréchet and Gâteaux derivatives agree. The converse is clearly not true, since the Gâteaux derivative may fail to be linear or continuous. In fact, it is even possible for the Gâteaux derivative to be linear and continuous but for the Fréchet derivative to fail to exist." Thus, i think even if the Gateaux derivative of functionals between complex Banach spaces is linear, we still can't guarantee the existence of F-derivative. $\endgroup$ – Paradiesvogel Jan 14 '18 at 6:15
  • $\begingroup$ Besides, @md2perpe, I will carefully read the paper written by Zorn (1946) then. Again, your kind help is much appreciated! :) $\endgroup$ – Paradiesvogel Jan 14 '18 at 6:22

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.