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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 $A = A^*$

Questions

  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.
Community
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alisianoi
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1 Answers1

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  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.

user127096
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