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Let $f \in \mathbb{C}[x],z\in\mathbb{C}$ with $f=\sum_{k=0}^nf_kx^k$ and $||f||_2=\sqrt{\sum_{k=0}^n |f_k|^2}$.

Prove:

$$\Vert (x-z)f\Vert_2=\Vert(z^*x-1)f\Vert_2$$ where $z^*$ denotes the complex conjugate. We may use the fact that $|w|^2=ww^*$.

For the left-hand side I have:

$$\begin{align}\Vert(x-z)f\Vert_2^2=&\sum_{k=0}^n|(x-z)f|^2=\sum_{k=0}^n(x-z)f(x-z)^*f^*=\sum_{k=0}^n (x-z)f(x^*-z^*)f^*\\=&\sum_{k=0}^n|f|^2(xx^*-zx^*-xz^*+zz^*)=\sum_{k=0}^n|f|^2(|x|^2-zx^*-xz^*+|z|^2)\end{align}$$

And for the right-hand side:

$$\begin{align}\Vert(z^*x-1)f\Vert_2^2=&\sum(z^*x-1)f(z^{**}x^*-1^*)f^*=\sum_{k=0}^n|f|^2(z^*x-1)(zx^*-1)\\=&\sum_{k=0}^n|f|^2(|z|^2|x|^2-zx^*-z^*x+1)\end{align}$$

Is this correct?

Comparing the terms in the parentheses after the $|f|^2$ we see that the the middle part cancels out but how can I complete the proof? Thanks for any answers!

mathtronaut
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1 Answers1

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Your proof is not correct, see my remarks below.

The calculation becomes relatively simple if one uses the scalar product in $\Bbb C[x]$ defined as $$ \langle f, g \rangle = \sum_k f_k g_k^* $$ for $f, g \in \Bbb C[x]$ with $f=\sum_{k} f_k x^k$, $g=\sum_{k} g_k x^k$. It is straightforward to verify that this satisfies all requirements of a (complex) scalar product.

Then $$ \Vert f \Vert_2 = \sqrt{\langle f, f \rangle} $$ and therefore $$ \begin{align} \Vert (x-z)f \Vert_2^2 &= \langle xf - zf, xf- zf \rangle \\ &= \langle xf, xf \rangle - \langle xf, zf \rangle - \langle zf, xf \rangle + \langle zf, zf \rangle \\ &= \Vert xf \Vert_2^2 - z^* \langle xf, f \rangle - z \langle f, xf \rangle + |z|^2 \Vert f \Vert_2^2 \end{align} $$ and $$ \begin{align} \Vert (z^*x-1)f \Vert_2^2 &= \langle z^*xf - f, z^*xf- f \rangle \\ &= \langle z^* xf, z^*xf \rangle - \langle z^*xf, f \rangle - \langle f, z^*f \rangle + \langle f, f \rangle \\ &= |z|^2\Vert xf \Vert_2^2 - z^* \langle xf, f \rangle - z \langle f, xf \rangle + \Vert f \Vert_2^2 \, . \end{align} $$ These expressions are equal because $\Vert xf \Vert_2 = \Vert f_2 \Vert$: the polynomials $f$ and $xf$ have the same nonzero coefficients, only at different positions.

Regarding your calculation: $(x-z)f$ is the polynomial $$ \sum_{k=0}^n (x-z) f_k x^k = -z f_0 + \sum_{k=1}^n (f_{k-1}-zf_k)x^k + f_n x^{n+1} $$ so that $$ \tag{$*$} \Vert (x-z)f \Vert_2^2 = |z f_0|^2 + \sum_{k=1}^n |f_{k-1}-zf_k|^2 + |f_n|^2 $$ and that is different from what you calculated for $\Vert (x-z)f \Vert_2^2$.

One can work with $(*)$ to verify the identity $\Vert (x-z)f\Vert_2=\Vert(z^*x-1)f\Vert_2$ and that is effectively the same calculation as what I wrote above. Using the scalar product just makes it simpler.

Martin R
  • 128,226