Bessel inequality strictly less

[jgthb]

Hi everyone

Today during problem session we had this seemingly simple exercise, but I just can't crack it:

We should give an example of an [tex]x \in \ell^2[/tex] with strict inequality in the Bessel inequality (that is an x for which [tex]\sum_{k=1}^\infty |<x,x_k>|^2 < ||x||^2[/tex], where [tex](x_k)[/tex] is an orthonormal basis). I have tried a few things, e.g. defining [tex]x_k[/tex] in the following way:

[tex]x_1 = (\frac{1}{\sqrt(2)},\frac{1}{\sqrt(2)},0,\ldots)[/tex]
[tex]x_2 = (\frac{-1}{\sqrt(2)},\frac{1}{\sqrt(2)},0,\ldots)[/tex]
[tex]x_k = (0,\ldots,0,1,0,\ldots)[/tex], for [tex]k \geq 3[/tex]

and defining x as [tex]x = (1,1,0,0,\ldots)[/tex],

but that doesn't seem to work. Does anyone have a better idea?

 

[Landau]

Bessel inequality (that is an x for which

[tex]\sum_{k=1}^\infty |<x,x_k>|^2 < ||x||^2,[/tex]

where [tex](x_k)[/tex] is an orthonormal basis).

Bessel's inequality concerns an orthonormal set (sequence). If that set is a basis, then equality holds: 'Parseval's Identity'.

That said, why wouldn't your example work? You have ||x||^2=2 and that sum of squared inner products equals 2/sqrt(2).

 

[jgthb]

set of course, not basis...thank you

but it doesn't work:

[tex]\sum_{k=1}^\infty |<x,x_k>|^2 =|<x,x_1>|^2=|\frac{2}{\sqrt(2)}|^2=2[/tex],

or am I just really bad at simple calculations today?

 

[Landau]

Yes, you are right; I forgot to square. But this should come as no surprise, because your x_k's form a basis! Why don't you just throw x_1 and x_2 away? What you're left with is of course still orthonormal:
(0,0,1,0,0,0,...)
(0,0,0,1,0,0,...)
(0,0,0,0,1,0,...)
etc., but not a basis any more.

Then the sum of square of innner products is just 0, and ||x||^2=2. Certainly, 0<2 :)

 

[jgthb]

of course! that's actually a bit embarrassing, that I didn't think of that :)

Thank you so much, Landau

 

[Landau]

Your're welcome :)