2

Let $f:H\to H,\, x\mapsto x/\sqrt{1+|x|^2}$ for $H$ a real Hilbert space. Calculate the derivative of $f$.

I want to confirm if my calculation is correct. Setting $g(x):=(1+(x|x))^{-1/2}$ we have that $f(x)=g(x)\cdot x$ where the dot means the scalar multiplication in $H$ (not the inner product, just the product of a real number by a vector). Then applying the product rule we find

$$\partial f(x)h=\partial g(x)h\cdot x+g(x)\cdot h,\quad x,h\in H$$

where $\partial g(x)h=-(1+|x|^2)^{-3/2}(x|h)$. Is that correct or there is something wrong? Thank you in advance.

Guy Fsone
  • 25,237
  • 1
    I compute the derivative directly to double check your result.

    Letting $g(r):=(1+r^2)^{-\frac12}$, we consider $x, h\in H$ and we compute $$ \left.\frac{\partial}{\partial\epsilon}\right|_{\epsilon=0} g(|x+\epsilon h|)(x+\epsilon h)= 2g'(|x|)\langle x| h\rangle x + g(|x|)h. $$This is the same result as yours, so I think it is correct.

     – Giuseppe Negro Oct 31 '17 at 19:12
  • thank you @Giuseppe, I wasnt totally sure about the "bilinearity" of the scalar multiplication. I found here a question about that. –  Oct 31 '17 at 19:14
  • You seem not appy with this answer below is there something wrong? – Guy Fsone Feb 02 '18 at 15:23

1 Answers1

2

Let $g(t) = (1+t)^{-1/2}$ and $h(x) =|x|^2$ Then, $$f(x) = x\cdot g(|x|^2) =I(x)\cdot g\circ h(x) $$ Hence, for $k\in H$ we have, $$df_x(k) ~~= dI_x(k) \cdot g\circ h(x) +I(x) \cdot d_x[g\circ h(x)](k) \\= k \cdot g\circ h(x) +x \cdot d_x[g\circ h(x)](h) $$

Given that, $dI_x(k) = k$. By chain rule we have $$ d_x[g\circ h(x)](k) = d_{h(x)}[g]\circ d_{x}[h(x)](k) = g'({h(x)})\cdot d_{x}[h(x)](k)$$

Now, $$g'(t) = -\frac12(1+t)^{-3/2}\implies g'({h(x)})=-\frac12(1+|x|^2)^{-3/2} $$

And it is easy to check that, $$d_{x}[h(x)](k) =\langle x,k\rangle +\langle k,x\rangle =2Re\langle x,k\rangle$$

Finally, $$\color{blue}{df_x(k) = k(1+|x|^2)^{-1/2} - xRe\langle x,k\rangle(1+|x|^2)^{-3/2}} $$

Guy Fsone
  • 25,237