Proving that $X$ is a Banach Space and $Y$ is Not

[mathmari]

Hey!

Let $ T> 0 $ be fixed.

We denote $ X = \{f \in C (\mathbb{R}): f (t) = f (t + T) \ \forall t \in \mathbb {R} \} $ and $ Y = \{f \in C^1 (\mathbb{R}): f (t) = f (t + T) \ \forall t \in \mathbb {R} \} $ be the spaces of the $ T $ periodic continuous and continuously differentiable functions respectively, both given the maximum norm $ \displaystyle {\| f \|_{\infty} = \max_{t \in \mathbb {R}} | f (t) |} $.

I want to show that $X$ is a Banach space, but not $Y$ as a subspace of $X$.

Hint: Show that $Y$ is not complete. For that, show that the sequence $f_n(x)=\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}$ is a Cauchy sequence in $Y$ but doesn't converge in $Y$.
First we have to show that $X$ is a Banach space, i.e., that $X$ is complete, or not?

It holds that $C(\mathbb{R})$ is a Banach space, or not? Then $f_n(x)\in X\subseteq C(\mathbb{R})$. So a Cauchy sequence $\{f_n\}$ of $C(\mathbb{R})$ converges to $f\in C(\mathbb{R})$. It is left to show that if $f_n$ is $T$-periodic then $f$ is $T$-periodic.
Suppose that $f$ is not $T$-periodic, then $f(t)\neq f(t+T)$ for $t\in \mathbb{R}$.
Since $f_n$ is $T$ periodic, we have that $f_n(t)=f_n(t+T)$. We take the limit $n\rightarrow\infty$ and get $f(t)=f(t+T)$, a contradiction.
That means that a Cauchy sequence $\{f_n\}$ of $X$ converges to $X$, and so $X$ is complete.

Is this correct? (Wondering) Then we want to show that $Y$ is not a Banach space, don't we?

Let $N\in \mathbb{N}$ and $n,m>N$.

We have to show that $$|f_n(x)-f_m(x)|=\left| \frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}-\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{m}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}\right |<\epsilon$$ for every $\epsilon>0$ or not?

Could you give me a hint how we could show that? (Wondering) Then we have that $$\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}=\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\left |\sin\left (\frac{2\pi}{T}x\right )\right |}=\frac{\left |\sin\left (\frac{2\pi}{T}x\right )\right |^2}{\left |\sin\left (\frac{2\pi}{T}x\right )\right |}=\left |\sin\left (\frac{2\pi}{T}x\right )\right |$$ Since the absolute value is not continuously differentiable in $\mathbb{R}$, since it is not differentiable at $0$, it follows that the limit of $f_n(x)$ is not in $Y$ and so $f_n(x)$ doesn't convereg in $Y$.

So, since not every Cauchy sequence of $Y$ is convergent in $Y$, it follows that $Y$ is not complete, right? (Wondering)

 

[I like Serena]

Hey mathmari! (Wave)

What you have is correct.
So $X$ is indeed complete and therefore a Banach space.
And $Y$ has a Cauchy sequence that shows it is not complete.
Good job! (Happy)

 

[mathmari]

What you have is correct.
So $X$ is indeed complete and therefore a Banach space.
And $Y$ has a Cauchy sequence that shows it is not complete.
Good job! (Happy)

Great! (Yes)

How could we justify that $C(\mathbb{R})$ is a Banach space? (Wondering)

Do we have to show that $\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}$ is a Cauchy sequence of $Y$ ?
If yes, how could we do that? (Wondering)

 

[I like Serena]

How could we justify that $C(\mathbb{R})$ is a Banach space?

$C(\mathbb{R})$ means that $\forall t\in \mathbb R: \lim_{h\to 0} f(t+h)=f(t)$.
Will that be true if $f$ is the absolute converging limit of a sequence of functions in $C(\mathbb R)$? (Wondering)

Do we have to show that $\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}$ is a Cauchy sequence of $Y$ ?

That's not related to $C(\mathbb R)$ is it?
Didn't you already show that that particular Cauchy sequence is not in $Y$, showing that $Y$ is not a Banach space? (Worried)

 

[mathmari]

$C(\mathbb{R})$ means that $\forall t\in \mathbb R: \lim_{h\to 0} f(t+h)=f(t)$.
Will that be true if $f$ is the absolute converging limit of a sequence of functions in $C(\mathbb R)$? (Wondering)

I got stuck right now. I haven't really understood that part. (Thinking)

That's not related to $C(\mathbb R)$ is it?
Didn't you already show that that particular Cauchy sequence is not in $Y$, showing that $Y$ is not a Banach space? (Worried)

I thought that the Cauchy sequence should be in $Y$ but its limit is not in $Y$. (Wondering)

 

[Opalg]

Do we have to show that $\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}$ is a Cauchy sequence of $Y$ ?
If yes, how could we do that? (Wondering)

$$|f_n(x)-f_m(x)|=\left| \frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}-\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{m}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}\right | = \left| \frac{\left(\frac1m - \frac1n\right)\sin^2\left (\frac{2\pi}{T}x\right )}{\left(\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right)\right|\right) \left(\frac{1}{m}+\left |\sin\left (\frac{2\pi}{T}x\right)\right|\right) }\right| \leqslant \left| \frac{\left(\frac1m - \frac1n\right)\sin^2\left (\frac{2\pi}{T}x\right )}{\left |\sin\left (\frac{2\pi}{T}x\right)\right|^2}\right| = \left|\tfrac1m - \tfrac1n\right|,$$ which can be made less than $\varepsilon$ for all sufficiently large $m$ and $n$, because the sequence $\left\{\frac1n\right\}$ is Cauchy in $\Bbb{R}$. What's more, the right side of the inequality $|f_n(x)-f_m(x)| < \left|\tfrac1m - \tfrac1n\right|$ is independent of $x$. That shows that the sequence $\{f_n(x)\}$ is uniformly Cauchy, in other words it is Cauchy for the $\|\,.\|_\infty$-norm. That is the norm in both $Y$ and $X$. So $\{f_n(x)\}$ is Cauchy in both $Y$ and $X$.

Since $X$ is complete, it follows that the sequence $\{f_n(x)\}$ converges to a limit in $X$. You have shown that the sequence converges (pointwise) to a non-differentiable function in $X$. But if a sequence has a limit in the $\|\,.\|_\infty$-norm, then that limit must coincide with the pointwise limit. Since that limit is not differentiable, it is not in $Y$. So the sequence does not converge in $Y$ and therefore $Y$ is not complete.

 

[mathmari]

$$|f_n(x)-f_m(x)|=\left| \frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}-\frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{m}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}\right | = \left| \frac{\left(\frac1m - \frac1n\right)\sin^2\left (\frac{2\pi}{T}x\right )}{\left(\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right)\right|\right) \left(\frac{1}{m}+\left |\sin\left (\frac{2\pi}{T}x\right)\right|\right) }\right| \leqslant \left| \frac{\left(\frac1m - \frac1n\right)\sin^2\left (\frac{2\pi}{T}x\right )}{\left |\sin\left (\frac{2\pi}{T}x\right)\right|^2}\right| = \left|\tfrac1m - \tfrac1n\right|,$$ which can be made less than $\varepsilon$ for all sufficiently large $m$ and $n$, because the sequence $\left\{\frac1n\right\}$ is Cauchy in $\Bbb{R}$. What's more, the right side of the inequality $|f_n(x)-f_m(x)| < \left|\tfrac1m - \tfrac1n\right|$ is independent of $x$. That shows that the sequence $\{f_n(x)\}$ is uniformly Cauchy, in other words it is Cauchy for the $\|\,.\|_\infty$-norm. That is the norm in both $Y$ and $X$. So $\{f_n(x)\}$ is Cauchy in both $Y$ and $X$.

Since $X$ is complete, it follows that the sequence $\{f_n(x)\}$ converges to a limit in $X$. You have shown that the sequence converges (pointwise) to a non-differentiable function in $X$. But if a sequence has a limit in the $\|\,.\|_\infty$-norm, then that limit must coincide with the pointwise limit. Since that limit is not differentiable, it is not in $Y$. So the sequence does not converge in $Y$ and therefore $Y$ is not complete.

I understand! Thank you very much! (Smile)