- #1

mathmari

Gold Member

MHB

- 5,049

- 7

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)

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)

Last edited by a moderator: