Let $f:L(E)\to L(E),$ $u\mapsto f(u)=u^3=u\circ u\circ u.$ Then, $f$ is differentiable and compute $f'(u)$

Question

Let $$f:L(E)\to L(E)$$ $$u\mapsto f(u)=u^3=u\circ u\circ u.$$ I'm interested in proving that $f$ is differentiable and computing $f'(u).$

Here is what I've done:

$$f(u+h)-f(u)=(u+h)^3-u^3$$ $$=3 u^2 h+3uh^2+h^3$$ $$=L(u) h+3uh^2+h^3$$ Then, $$||\varepsilon (h)||=\frac{||3uh^2+h^3||}{||h||}$$ My question: I'm I on track? If yes, how do I show that $||\varepsilon (h)||\to 0$ as $h\to 0$? If no, can anyone show me a reference or a proof?

Answer 1

Our OP Mike seems to have grasped the essential idea, but it must be remembered that, since $L(E)$ does not form a commutative algebra, we must take care to preserve the order of products when expanding expressions such as $(u + w)^n$; this is illustrated below for the function

$f(u) = u^3, \tag 1$

we have

$f(u + h) = (u + h)^3 \tag 2$

for any $h \in L(E); \tag 3$

we may expand the right-hand side as follows:

$f(u + h) = (u + h)^3 = (u + h)(u + h)(u + h) = (u + h)(u^2 + uh + hu + h^2)$ $= u^3 + u^2 h + uhu + uh^2 + hu^2 + huh + h^2 u + h^3; \tag 4$

thus

$f(u + h) - f(u) = u^2 h + uhu + uh^2 + hu^2 + huh + h^2 u + h^3$ $= u^2 h + uhu + hu^2 + uh^2 + huh + h^2 u + h^3; \tag 5$

we note that each of the first three terms on the right of (5) is linear in $h$; thus we set,

$D_u f(h) = u^2 h + uhu + hu^2; \tag 6$

as for the last four terms, we find

$\Vert uh^2 + huh + h^2 u + h^3 \Vert \le \Vert uh^2 \Vert + \Vert huh \Vert + \Vert h^2 u \Vert + \Vert h^3 \Vert$ $\le 3\Vert u \Vert \Vert h \Vert^2 + \Vert h \Vert^3 = (3\Vert u \Vert + \Vert h \Vert) \Vert h \Vert^2 ; \tag 7$

we may assemble (5), (6) and (7) together to obtain

$\Vert f(u + h) - f(u) - D_u f(h) \Vert \le (3\Vert u \Vert + \Vert h \Vert) \Vert h \Vert^2, \tag 8$

whence

$\dfrac{\Vert f(u + h) - f(u) - D_u f(h) \Vert}{\Vert h \Vert} \le (3\Vert u \Vert + \Vert h \Vert) \Vert h \Vert; \tag 9$

therefore,

$\lim_{\Vert h \Vert \to 0} \dfrac{\Vert f(u + h) - f(u) - D_u f(h) \Vert}{\Vert h \Vert} = 0, \tag{10}$

which shows that our expression

$D_u f(h) = u^2 h + uhu + hu^2 \tag{11}$

itself a linear mapping $L(E) \to L(E)$, is the derivative of $f(u) = u^3$.

The above also shows how the error term $(3 \Vert u \Vert + \Vert h \Vert) \Vert h \Vert^2$ may be handled.

This result may be extended to any $g(u) = u^k$; for $D_ug(h)$ we obtain a sum of terms of the form $u^ihu^{k - i - 1}$ for $0 \le i \le k - 1$.