Let $ C_\infty$ be inner product space of all real sequences $\{x_n\}$ with $x_n$ finite number of nonzero terms and the inner product defined by

$$\langle x,y\rangle =\sum_{i=0}^\infty x_ny_n$$

I want to prove this is not a Hilbert space but I do not know how to apply Parallelogram law here.

So any help appreciated

  • $\begingroup$ I changed $C_{00}$ to $C_\infty$, {$x_n$} to $\{x_n\}$, $<x,y>$ to $\langle x,y\rangle$, and the now-displayed equality to a displayed (rather than inline) equality. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 15 '14 at 4:54

This is an inner product space and accordingly it satisfies the parallelogram law. Hence you cannot show that it is not a Hilbert space be showing it fails to satisfy the parallelogram law.

The reason it is not a Hilbert space is that it is not complete: there are some Cauchy sequences in it that fail to converge. For example, consider the sequence whose $n$th term is $$ (1,\ 1/2,\ 1/3,\ 1/4,\ \ldots,\ 1/n,\ 0,\ 0,\ 0,\ 0,\ \ldots). $$ There is no element of this space to which this converges. (I leave it as an excercise to show that it is a Cauchy sequence, but if you ask about that I might say more.)

  • $\begingroup$ how can I prove this sequence does not converge?thanks in advance @MichaelHardy $\endgroup$ – SSH Dec 15 '14 at 5:27
  • $\begingroup$ First show that the function that maps the sequence to its $n$th component is continuous. Use that to conclude that the $n$th component of the limiting vector would have to be $1/n$. Then show (this part is easy) that there is no vector in this space whose $n$th component is $1/n$ for every $n$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 15 '14 at 6:07

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