4

Let $I$ be a set, and let $B(I)$ be the space of bounded, real-valued functions on $I$ equipped with the sup-norm. Let $\phi: \mathbb R \to \mathbb R$ be bounded and continuously differentiable everywhere. Finally, let $S: B(I) \to \mathbb R$ be linear and continuous.

Define $\Phi: B(I) \to \mathbb R$ by $\Phi(x) = S(\phi \circ x)$.

Is $\Phi$ Fréchet differentiable at every $x \in B(I)$, and if so is it the case that $\Phi'(x) = S(\phi' \circ x)$?

I can show that the function $x \mapsto S(\phi' \circ x)$ is linear and bounded, using the corresponding facts about $\phi'$ and $S$, but I'm not sure I can show that this function satisfies the definition of the Fréchet derivative.

I have to show, for every $x \in B(I)$, that $$\lim_{\| h \|_\infty \to 0} \frac{| \Phi (x + h) - \Phi(x) - S(\phi' \circ h) |}{\| h \|_\infty} = 0,\tag{1}$$ where $\| \cdot \|_\infty$ is the sup-norm on $B(I)$. Now, by the definition of $\Phi$ and the linearity of $S$ $$ \frac{\Phi (x + h) - \Phi(x) - S(\phi' \circ h)}{\|h\|_\infty} = \frac{S(\phi \circ (x+h)) - S(\phi \circ x) - S(\phi' \circ h)}{\|h\|_\infty} = S\Big(\frac{[\phi \circ(x+h)] - [\phi \circ x] - [\phi' \circ h]}{\|h\|_\infty} \Big).$$ From here I would like to argue that as $\|h\|_\infty \to 0$, $$\frac{[\phi \circ(x+h)] - [\phi \circ x] - [\phi' \circ h]}{\|h\|_\infty} \to 0, \tag{2}$$ and then use the continuity of $S$ to conclude. I haven't convinced myself that (2) holds however.

aduh
  • 8,662
  • bounded convex functions are in fact constant, so in this sense the question is rather "trivial". But as long as $\phi'$ is uniformly continuous, you can show that $(2)$ goes to $0$ (just evaluate the numerator at a point $t\in I$, write a first order Taylor expansion with Lagrange remainder, and then use uniform continuity of $\phi'$ to show that it can be made sufficiently small) – peek-a-boo Aug 09 '20 at 00:40
  • @peek-a-boo Right. I don't think the convexity is really relevant, so I will just drop it. – aduh Aug 09 '20 at 00:50

1 Answers1

2

For every $t\in I$, by the differentiability of $\phi$, $$(\phi (x+h))(t) - \phi(x(t))=\phi(x(t) + h(t)) - \phi(x(t)) =\phi'(x(t))h(t) + o(h(t)).$$ That is to say, at least formally, $$ \phi\circ (x+h) - \phi\circ x = (\phi'\circ x) h + o(h).$$ To justify this, we need to use $\phi\in C^1$ to write for $x,h\in \mathbb R$, $$ \phi(x+h) - \phi(x) - \phi'(x)h= R(x,h)h$$ and the boundedness of $\phi'$ gives the boundedness of $R$ on bounded subsets on $\mathbb R^2$. Letting $x,h\in B(I)$ gives the result.

By chain rule and linearity, $d (S\circ f)(x)h = dS(f(x))[df(x)h]=S(df(x)h)$, so $$ d\Phi(x)h = S((\phi'\circ x) h). $$ Note carefully the difference between $h\mapsto (\phi'\circ x) h$ and $x\mapsto \phi'\circ x$. The Frechet derivative at each point is supposed to be a linear map, and I don't think you could have shown $x\mapsto S(\phi'\circ x)$ is linear. Well, actually you could, because the only convex, bounded functions $\phi$ are constants, and so $\Phi$ is a constant function, with 0 derivative, and abusing notation with 0 would give you linearity. But ignoring either boundedness or convexity(as you have edited the question), your formula doesn't give a linear map in the general case. The map $x\mapsto S((\phi'\circ x)\cdot)$ maps from $B(I)$ to the space of linear maps on $B(I)$ taking values in $\mathbb R$.

Calvin Khor
  • 34,903
  • Thanks. You're right, I miswrote the thing about linearity. – aduh Aug 09 '20 at 00:53
  • What is $f$ supposed to be? – aduh Aug 09 '20 at 00:54
  • $f$ is an arbitrary function with the correct domain/codomain from the statement of chain rule I had floating in my head @grndl if that's not clear I can attempt to reword – Calvin Khor Aug 09 '20 at 00:54
  • @grndl well actually I guess $f=\phi$ but as a function $B(I)\to B(I)$ – Calvin Khor Aug 09 '20 at 00:56
  • Just to clarify, in the chain rule calculation, are you using the fact that $S$ is differentiable because it's linear and continuous, and, moreover, $dS(x) = S$ for all $x$? – aduh Aug 09 '20 at 01:00
  • @grndl yes, exactly – Calvin Khor Aug 09 '20 at 01:01
  • I don't think going from the first step to the second (i.e from $o(h(t))$ to $o(h)$) is as obvious as you make it seem, because the first is a pointwise matter which follows right from the definition of differentiability of $\phi$, while the second requires a uniform approximation (atleast I don't see any way to weaken it) for which I had to invoke continuous differentiability of $\phi$, boundedness of $x(\cdot)$, and also use uniform continuity of the remainder term, $R$, in $\phi(\xi+\eta) = \phi(\xi) + \phi'(\xi)\cdot \eta + R(\xi,\eta)\cdot \eta$ on compact subsets of $\Bbb{R}^2$. – peek-a-boo Aug 09 '20 at 01:46
  • @peek-a-boo yes, you're right, I glossed over that. I'll edit that in – Calvin Khor Aug 09 '20 at 01:57