Differentiability, linear operators

Question

Let $Y$ be a complete normed linear space, and let $M$ denote the space of bounded linear operators from $Y$ to itself. Let $L : M → M$ be the map defined by $L(A) := A^2$.

I am supposed to show that L is differentiable at each $A\in M$, and then find the derivatives.

Tried to find the derivatives like this:

$L'(A) = \lim_{h \to 0} \frac{L(A+h)-L(A)}{h}= \lim_{h \to 0} \frac{(A+h)^2-A^2}{h}= \lim_{h \to 0} \frac{A^2+2Ah-h^2-A^2}{h} = \lim_{h \to 0} 2A+h = \underline{2A}$

If this is correct, all I have to do now is to show that $L$ is differentiable at all $A\in M$, and I'm wondering if I can do that by saying that $L$ is a composition of $A$ and $A$ since $A$ is a linear operator, and all linear operators are differentiable?

Answer 1

The derivative of $L$ at an operator $A$ is a linear map. In this case, it is the map $\varphi$ defined by $\varphi(M)=AM+MA$. In fact\begin{align}\lim_{M\to A}\frac{\bigl\lVert L(M)-L(A)-\varphi(M-A)\bigr\rVert}{\lVert M-A\rVert}&=\lim_{M\to A}\frac{\bigl\lVert M^2-A^2-A(M-A)-(M-A)A\bigr\rVert}{\lVert M-A\rVert}\\&=\lim_{M\to A}\frac{\lVert M^2-AM-MA+A^2\rVert}{\lVert M-A\rVert}\\&=\lim_{M\to A}\frac{\bigl\lVert(M-A)^2\bigr\rVert}{\lVert M-A\rVert}\\&=0.\end{align}

Answer 2

In order to discover (and at the same time prove that $D_A(H):=AH+HA$ is the differential of function $L$, which is different from the derivative), just expand

$$L(A+H)=(A+H)^2=\underbrace{A^2}_{L(A)}+\underbrace{(AH+HA)}_{D_A(H)}+\underbrace{H^2}_{2nd \ order}$$

and collect all the first degree terms.

Do you see the connection with Taylor expansion :

$$f(a+h)=f(a)+\underbrace{f'(a).h}_{differential}+\text{terms in} \ h^2, h^3, \text{etc.} ?$$