Problem statement

Let $x \in l^2$ and $J(x) = \sum_{n = 1}^{+\infty} x_{2n - 1}^2$

Find first and second Frechet derivatives.

Attempted solution

Let's note that $J(x) = \sum_{n = 1}^{+\infty}x^2_{2n-1} = \left<Ax, x\right>$ where operator $A$ does the following: it takes a vector and substitutes all of its even coordinates with $0$. Then the derivatives look like:

$$ J(x + h) - J(x) = \left<A(x + h), h\right> - \left<A, x\right> = \left<(A + A^*)x, h\right> + \left<Ah, h\right> $$

This means that $DJ(x)(h) = \left<(A + A^*)x, h\right>$ and $D^2J(x)(h)$ will look like this:

$$ J(x + h + k) - J(x) - DJ(x)(h + k) = \left<A(x + h + k), x + h + k\right> - \left<Ax, x\right> + \left<(A + A^*)x, h + k\right> = \left<A(h + k), h + k\right> $$

Now, to write down the answer, I suppose I need to understand what $A^*$ actually is. That proved to be quite hard, following this question I tried from the definition:

$\left<Ax, y\right> = \left<x, A^*y\right>$ implies $\left<Ax, x\right> = \left<x, A^*x\right> = J(x)$ which means $A = A^*$


  1. Are there any mistakes in the derivatives calculation?
  2. Is that really true that $A = A^*$? It must act the same then?
  3. I know that second-order Frechet derivatives are isomorphic to bilinear forms. However, I do not understand where this first line comes from $J(x + h + k) - J(x) - DJ(x)(h + k)$ I borrowed that from a previous question of mine.
  1. Your computation is correct

  2. Yes, $A$ is self-adjoint. More generally, if $(\mu_n)$ is any bounded sequence of real numbers, then the multiplication operator $(x_n) \mapsto (\mu_n x_n)$ is self-adjoint. This is the infinite-dimensional analogue of diagonal matrix with real entries. In your example, $\mu_n$ alternates between $0$ and $1$. Put more geometrically, $A$ is the orthogonal projection onto the space of sequences supported on odd indices only. Note that $J$ is the square of the norm of said projection.

  3. If you are asking where $J(x + h + k) - J(x) - DJ(x)(h + k)$ comes from: I understand this is not the definition of derivative, but if $D^2J$ exists, then it can be calculated with this simple formula. If you want a justification based on the definition, please specify the definition you are using.


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.