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My main problem is that I do not know how to show that the orthonormal set $\{g_k\}_{k=1}^{\infty}$ is incomplete on $R\bigl([0;\pi]\bigr)$ which is the set of the integrable function on $[0;\pi]$. $$ \{g_k\}_{k=1}^{\infty}:=\left\{\sqrt{ \frac{2}{\pi} }\cos(2kx)\right\}_{k=1}^{\infty} $$ We define the scalar product as: $\langle f,g\rangle =\int_{a}^{b}\bar{f}(x)g(x)\,dx$

The completeness of a set $\{f_k\}_{k=0}^{\infty}$ on a set $A$. $A:=\{f\mid f \;\text{meets certain condition on the interval}\; [a;b]\}$ $$\lim_{n\to\infty} \Bigl\lVert f-\sum_{k=1}^{n} c_k f_k\Bigr\rVert=\lim_{n\to\infty}\sqrt{\Bigl\langle f-\sum_{k=1}^{n} c_k f_k , f-\sum_{k=1}^{n} c_k f_k\Bigr\rangle}=0$$ $$\text{where} \quad c_k:=\langle f,f_k\rangle $$

When I go straightforward, I get the following. $$ \begin{split} \Bigl\langle f-\sum_{k=1}^{n} d_k g_k , f-\sum_{k=1}^{n} d_k g_k\Bigr\rangle & =\langle f,f\mkern1.5mu\rangle +\Bigl\langle \sum_{k=1}^{n} d_k g_k,\sum_{k=1}^{n} d_k g_k\Bigr\rangle-\Bigl(\Bigl\langle f,\sum_{k=1}^{n} d_kg_k\Bigr\rangle+\Bigl\langle\sum_{k=1}^{n} d_k g_k,f\Bigr\rangle\Bigr)\\ & =||f||^2+\sum_{k,j=1}^{n} \bar{d_j}d_k\langle g_j,g_k\rangle-\biggl(\Bigl\langle f,\sum_{k=1}^{n} d_kg_k\Bigr\rangle +\overline{\Bigl\langle f,\sum_{k=1}^{n} d_kg_k\Bigr\rangle} \biggr)\\ & =||f||^2+\sum_{k=1}^{n} |d_k|^2-2\Re\Bigl(\Bigl\langle f,\sum_{k=1}^{n} d_kg_k\Bigr\rangle\Bigr)\\ & =||f||^2+\sum_{k=1}^{n} |d_k|^2-2\Re\Bigl(\sum_{k=1}^{n}d_k\langle f,g_k\rangle\Bigr)\\ & =||f||^2+\sum_{k=1}^{n} |d_k|^2-2\Re\Bigl(\sum_{k=1}^{n}d_k^2\Bigr)\\ \end{split} $$ At this point I don't really know where to go, and if I'm following the right path to the demonstration.

Maybe I should simply give a counter-example to prove it, but I have no clue what exemple would show that the set $\{g_k\}_{k=1}^{\infty}$ is incomplet.

I would still prefer to have a nice rigourous proof to show the assumption.

Thank you in advance for your responses.

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    $\begingroup$ If you find some function $f$ such that $(f,g_k)=0$ for all $k$ and $f\neq 0$, then $g_k$ is not complete, because your computations shows that $\|f-\sum a_kg_k\|^2\geq \|f\|^2$. Try $f(x)=\sin(2x)$. $\endgroup$ – user552631 Apr 15 '18 at 12:36
  • $\begingroup$ @jyre Allright, thank you $\endgroup$ – Beginner Apr 15 '18 at 12:53
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Note that $$ \forall k\in\mathbb{N^*} \quad \langle \cos(2kx),1\rangle = \left.\int_{0}^{\pi}\cos(2kx)dx=\frac{\sin(2kx)}{2k}\right|_{x=0}^{\pi} = 0 $$ So the constant function $1$ is orthogonal to all of the functions in your set, which proves that your orthonormal set is not complete.

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If we define that $d_k=a_k+ib_k$, where $a_k,b_k\in\mathbb{R}$. $$ \begin{split} \sum_{k=1}^{n}|d_k|^2-2\Re\left(\sum_{k=1}^{n}(d_k)^2\right) & =\sum_{k=1}^{n} a_k^2+b_k^2 -2\Re\left(\sum_{k=1}^{n}a_k^2-b_k^2+i(2a_kb_k)\right) \\ & =\sum_{k=1}^{n} a_k^2+b_k^2 -2\sum_{k=1}^{n}a_k^2-b_k^2 \\ & =\sum_{k=1}^n -a_k^2+3b_k^2 \end{split} $$ From this point it is hard to conclude something.

Let's only give an example, that shows that the set is not complete.

Suppose that f(x)=sin(2x). $$ d_k=<f,g_k>=0 $$ Therefore we have that $$ ||f-\sum_{k=1}^nd_kg_k||=||f||=\frac{\pi}{2}\neq0 $$ Which means that the set is not complete.

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