Show that a vector space is not complete (therefore not a Hilbert spac

[fluidistic]

Homework Statement

Consider the space of continuous functions in [0,1] (that is C([0,1]) over the complex numbers with the following scalar product: $\langle f , g \rangle = \int _0 ^1 \overline{f(x)}g(x)dx$.
Show that this space is not complete and therefore is not a Hilbert space.
Hint:Find a Cauchy sequence (with respect to the norm $||f||=\sqrt {\langle f,f \rangle }$ of functions in C([0,1])) such that the sequence converges to a non continuous function.

Homework Equations

Hmm.

The Attempt at a Solution

So I have in mind a sequence that reprent a truncated Fourier series that would represent a square wave if the series is never truncated. A kind of Heaviside function.
Or a sequence of Gaussians that would reprent the Dirac delta (but it's not a function).
In that case $f_n=Ce^{-ax^2n^2}$ would make it? I'm somehow confused when n tends to infinity to what happens at x=0.

 

[Dick]

Homework Statement

Consider the space of continuous functions in [0,1] (that is C([0,1]) over the complex numbers with the following scalar product: $\langle f , g \rangle = \int _0 ^1 \overline{f(x)}g(x)dx$.
Show that this space is not complete and therefore is not a Hilbert space.
Hint:Find a Cauchy sequence (with respect to the norm $||f||=\sqrt {\langle f,f \rangle }$ of functions in C([0,1])) such that the sequence converges to a non continuous function.

Homework Equations

Hmm.

The Attempt at a Solution

So I have in mind a sequence that reprent a truncated Fourier series that would represent a square wave if the series is never truncated. A kind of Heaviside function.
Or a sequence of Gaussians that would reprent the Dirac delta (but it's not a function).
In that case $f_n=Ce^{-ax^2n^2}$ would make it? I'm somehow confused when n tends to infinity to what happens at x=0.

Try something much simpler, like $f_n=x^n$. What does that do?

 

[fluidistic]

Try something much simpler, like $f_n=x^n$. What does that do?

Hmm since x is between 0 and 1, the sequence would converge to 0.
Edit: Ah yeah at x=1 it's worth 1, a discontinuous function. Wow.