Is that subset of the set of continuous differential functions closed?

[approx1mate]

Hi! I have used the physics forum a lot of times to deal with several tasks that I had and now its the time to introduce my own query! So please bear with me :-)

Homework Statement

Equip the set C^1_{[0,1]} with the inner product:
<br /> \left\langle f,g \right\rangle= \int_{0}^{1} f(x)\bar{g(x)} + \int_{0}^{1} f&#039;(x)\bar{g&#039;(x)}dx<br />
(the bar above the g function is the conjugate symbol)
I need to show that the subspace:

<br /> W = \{f\in C^1_{[0,1]} | f(1)=0\}<br />

is a closed subspace of C^1_{[0,1]}.

Homework Equations

\left\langle f,cosh \right\rangle = f(1)sinh(1).

The Cauchy inequality: |\left\langle f,g \right\rangle | \le \|f\|\|g\|,
the Pythagoras theorem: \|f+g\|^2 = \|f\|^2 + \|g\|^2,
the parallelogram law: \|f+g\|^2 + \|f-g\|^2 = 2(\|f\|^2 + \|g\|^2),
the triangular inequality: \|f+g\| \le \|f\| + \|g\|

The Attempt at a Solution

Let us take a Cauchy sequence \{f^n\}_{n=1}^{\infty} \in W, because
(C^1_{[0,1]},\|\cdot \|) is a Hilbert space then the sequence \{f^n\}_{n=1}^{\infty} converges to f\in C^1_{[0,1]}.
Therefore it only remains to be shown that at the limit f(1)=0.

At this point I am stuck. I can see that the cosh function is orthogonal
to the set W and I also tried to use the above "relevant equations" but
I couldn't see what would be a possible proof.

Any advice?

 

[micromass]

Hi! I have used the physics forum a lot of times to deal with several tasks that I had and now its the time to introduce my own query! So please bear with me :-)

Homework Statement

Equip the set C^1_{[0,1]} with the inner product:
<br /> \left\langle f,g \right\rangle= \int_{0}^{1} f(x)\bar{g(x)} + \int_{0}^{1} f&#039;(x)\bar{g&#039;(x)}dx<br />
(the bar above the g function is the conjugate symbol)
I need to show that the subspace:

<br /> W = \{f\in C^1_{[0,1]} | f(1)=0\}<br />

is a closed subspace of C^1_{[0,1]}.

Homework Equations

\left\langle f,cosh \right\rangle = f(1)sinh(1).

The Cauchy inequality: |\left\langle f,g \right\rangle | \le \|f\|\|g\|,
the Pythagoras theorem: \|f+g\|^2 = \|f\|^2 + \|g\|^2,
the parallelogram law: \|f+g\|^2 + \|f-g\|^2 = 2(\|f\|^2 + \|g\|^2),
the triangular inequality: \|f+g\| \le \|f\| + \|g\|

The Attempt at a Solution

Let us take a Cauchy sequence \{f^n\}_{n=1}^{\infty} \in W, because
(C^1_{[0,1]},\|\cdot \|) is a Hilbert space then the sequence \{f^n\}_{n=1}^{\infty} converges to f\in C^1_{[0,1]}.
Therefore it only remains to be shown that at the limit f(1)=0.

At this point I am stuck. I can see that the cosh function is orthogonal
to the set W and I also tried to use the above "relevant equations" but
I couldn't see what would be a possible proof.

Any advice?

First of all, I very much doubt that your space is a Hilbert space. It has an inner-product, but are you sure it is complete. Did you prove it??

Anyway, you don't really need it for the proof. For the proof you must take a convergent sequence (f_n)_n in W. So you know that f_n(1)=0 for all n. You must prove that the limit f also has f(1)=0.

You must not prove that (f_n)_n is Cauchy and you must not prove that it converges. You know convergence from the hypothesis.

 

[approx1mate]

First of all, I very much doubt that your space is a Hilbert space. It has an inner-product, but are you sure it is complete. Did you prove it??

You are right, I was wrong about that.

Anyway, you don't really need it for the proof. For the proof you must take a convergent sequence (f_n)_n in W. So you know that f_n(1)=0 for all n. You must prove that the limit f also has f(1)=0.

Ok, that was my thought from the very beginning. But if I need just the f(1)=0, then how do I know that the limit f is also a continuous differentiable function? Don't I need that as well?

 

[micromass]

You are right, I was wrong about that.



Ok, that was my thought from the very beginning. But if I need just the f_n(1)=0, then how do I know that the limit f is also a continuous differentiable function? Don't I need that as well?

No, you have that.

So what you have is that a sequence (f_n)_n in W converges to a function f\in C^1_{[0,1]}. You must show f to be in W.

 

[approx1mate]

No, you have that.

So what you have is that a sequence (f_n)_n in W converges to a function f\in C^1_{[0,1]}. You must show f to be in W.

Oh yes, I am sorry, all that time I didn't pay any attention to the definition of W.
Thanks

 

[approx1mate]

I think that the solution is:

|\left\langle f^n-f,cosh \right\rangle| \le \|f^n-f\|\|cosh\|

But \forall \epsilon \ \ \exists n_o(\epsilon) such that for any
n \ge n_0(\epsilon) we have that:

|\left\langle f^n-f,cosh \right\rangle| \le \epsilon \|cosh\|

Therefore \left\langle f^n-f,cosh \right\rangle \to 0 \Leftrightarrow \left\langle f^n,cosh \right\rangle - \left\langle f,cosh \right\rangle \to 0 \Leftrightarrow

f(1)sinh(1) \to 0 \Leftrightarrow f(1)\to 0

 

[micromass]

Looks ok. However in the last line, you don't want =0 but rather \rightarrow 0.

So you got

&lt;f^n-f,cosh&gt;\rightarrow 0

for example.

 

[approx1mate]

Looks ok. However in the last line, you don't want =0 but rather \rightarrow 0.

So you got

&lt;f^n-f,cosh&gt;\rightarrow 0

for example.

Yes you are right again, thanks!