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:
$$\left\langle f,g \right\rangle= \int_{0}^{1} f(x)\bar{g(x)} + \int_{0}^{1} f'(x)\bar{g'(x)}dx$$
(the bar above the $g$ function is the conjugate symbol)
I need to show that the subspace:

$$W = \{f\in C^1_{[0,1]} | f(1)=0\}$$

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:
$$\left\langle f,g \right\rangle= \int_{0}^{1} f(x)\bar{g(x)} + \int_{0}^{1} f'(x)\bar{g'(x)}dx$$
(the bar above the $g$ function is the conjugate symbol)
I need to show that the subspace:

$$W = \{f\in C^1_{[0,1]} | f(1)=0\}$$

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

$$<f^n-f,cosh>\rightarrow 0$$

for example.

 

[approx1mate]

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

So you got

$$<f^n-f,cosh>\rightarrow 0$$

for example.

Yes you are right again, thanks!