Tanh is uniformly continous on R

[mathmari]

Hey!

We consider the function $f(x)=\tanh x$.

I shown so far the following:

• The function is defined for all $x\in \mathbb{R}$, i.e. $D_f=\mathbb{R}$.
• The function is strictly increasing.
• $\displaystyle{\lim_{x\rightarrow -\infty}\tanh x=-1}$ and $\displaystyle{\lim_{x\rightarrow \infty}\tanh x=1}$. And since $\tanh$ is continuous on the whole $\mathbb{R}$ the range is $(-1,1)$.

Now I want to show that $\tanh$ is uniformly continuous on $\mathbb{R}$ (without using differential calculus).

Could you give me a hint how we could show that? Do we have to use the definition? (Wondering)

 

[Euge]

Let $0 < \epsilon < 1$ and choose $\delta = \tanh^{-1}(0.5\epsilon)$. Then $\delta > 0$. Show that for all $x,y\in \Bbb R$, $\lvert x - y\rvert < \delta$ implies $\lvert \tanh(x) - \tanh(y)\rvert < \epsilon$, using the identity
$$\tanh(x) - \tanh(y) = [1 - \tanh(x)\tanh(y)]\tanh(x-y)$$
and the fact that the hyperbolic tangent is bounded by $1$.

 

[mathmari]

Let $0 < \epsilon < 1$ and choose $\delta = \tanh^{-1}(0.5\epsilon)$. Then $\delta > 0$. Show that for all $x,y\in \Bbb R$, $\lvert x - y\rvert < \delta$ implies $\lvert \tanh(x) - \tanh(y)\rvert < \epsilon$, using the identity
$$\tanh(x) - \tanh(y) = [1 - \tanh(x)\tanh(y)]\tanh(x-y)$$
and the fact that the hyperbolic tangent is bounded by $1$.

We have the following:
\begin{align*}|\tanh(x) - \tanh(y)| &= |[1 - \tanh(x)\tanh(y)]\tanh(x-y)| \\ & =|[1 - \tanh(x)\tanh(y)]|\cdot |\tanh(x-y)|\\ & \leq \left (|1| + |\tanh(x)\tanh(y)|\right )\cdot |\tanh(x-y)| \\ & \leq \left (1 + 1\right )\cdot |\tanh(x-y)| \\ & = 2 |\tanh(x-y)|\end{align*}
We have that $\tanh$ is increasing and that $x-y\leq |x-y|$ and that $|x-y|<\delta$. So we get that \begin{align*}|\tanh(x) - \tanh(y)|&\leq 2 |\tanh(x-y)| \\ & \leq 2 |\tanh|x-y||\\ & < 2 |\tanh(\delta)| \\ & = 2 |\tanh(\tanh^{-1}(0.5\epsilon))| \\ & = 2|0.5\epsilon|\\ & = 2\cdot 0.5\epsilon \\ & = \epsilon\end{align*}
Is everything correct? (Wondering) Is this the only way to show the uniform continuity? Because at the next subquestion we have to prove the addition formula, and if we use the definition we use that addition formula. (Wondering)

 

[Euge]

Is this the only way to show the uniform continuity? Because at the next subquestion we have to prove the addition formula, and if we use the definition we use that addition formula. (Wondering)

Here is another way. The function $f$ is continuous on $[0, \infty)$ with limit $1$ at infinity, so it is uniformly continuous on $[0, \infty)$. Since $f$ is continuous on $(-\infty,0]$ and $\lim\limits_{x\to\infty} f(x) = -1$, then $f$ is uniformly continuous on $(-\infty, 0]$. Therefore, $f$ is uniformly continuous on $\Bbb R$.

 

[mathmari]

Here is another way. The function $f$ is continuous on $[0, \infty)$ with limit $1$ at infinity, so it is uniformly continuous on $[0, \infty)$. Since $f$ is continuous on $(-\infty,0]$ and $\lim\limits_{x\to\infty} f(x) = -1$, then $f$ is uniformly continuous on $(-\infty, 0]$. Therefore, $f$ is uniformly continuous on $\Bbb R$.

Ah ok!

So, in general does it hold that if a function is continuous on $[0, \infty)$ (respectively on $(-\infty, 0]$ ) with a finit limit at infinity then the function is on that interval uniformly continous? (Wondering)

 

[I like Serena]

Ah ok!

So, in general does it hold that if a function is continuous on $[0, \infty)$ (respectively on $(-\infty, 0]$ ) with a finit limit at infinity then the function is on that interval uniformly continous?

It's listed in the properties section on wiki. (Angel)

 

[mathmari]

It's listed in the properties section on wiki. (Angel)

Ah ok!

To prove this, we have the following:

Suppose that $f:[a, \infty)\rightarrow \mathbb{R}$ is continuous and that $\displaystyle{\lim_{x\rightarrow \infty}f(x)=\ell}$.

Let $\epsilon >0$.

Since $\displaystyle{\lim_{x\rightarrow \infty}f(x)=\ell}$ there is a $c>a$ such that $|f(x)-\ell|<\frac{\epsilon}{3}$ for each $x\geq c$.

We have that $f$ is continuous in $[a, \infty)$, so $f$ is continuous on the closed interval $[a,c]$, that means that on this interval the function is uniformly continuous. So, there is a $\delta>0$ such that for each $x,y\in [a,c]$ with $|x-y|<\delta$ we have that $|f(x)-f(y)|<\frac{\epsilon}{3}$.

For each $x,y\in [a,\infty)$ with $|x-y|<\delta$ we have the following cases:

• $x,y\leq c$:
  
  We have that $|f(x)-f(y)|<\frac{\epsilon}{3}<\epsilon$.
  
• $x,y\geq c$:
  
  We have that \begin{align*}|f(x)-f(y)|&=|f(x)-\ell+\ell-f(y)|=|(f(x)-\ell)-(f(y)-\ell)|\leq |f(x)-\ell|+|f(y)-\ell| \\ & <\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2\epsilon}{3}<\epsilon\end{align*}
  
• $x\leq c\leq y$:
  
  What could we do in this case? (Wondering)
  

Are the first two cases correct? (Wondering)

 

[Euge]

What could we do in this case? (Wondering)

Start by writing $\lvert f(x) - f(y)\rvert \le \lvert f(x) - f(c)\rvert + \lvert f(c) - \ell \rvert + \lvert \ell - f(y)\rvert$.

Are the first two cases correct? (Wondering)

Yes, they're correct.

 

[mathmari]

Start by writing $\lvert f(x) - f(y)\rvert \le \lvert f(x) - f(c)\rvert + \lvert f(c) - \ell \rvert + \lvert \ell - f(y)\rvert$.

We have that $x\leq c$. For each $x,y\in [a,c]$ with $|x-y|<\delta$ it follows that $\lvert f(x) - f(y)\rvert<\frac{\epsilon}{3}$. Does it hold that $|x-c|<\delta$ ? (Wondering)

Since $c\geq c$ we have that $\lvert f(c) - \ell \rvert<\frac{\epsilon}{3}$.

Since $y\geq c$ we have that $\lvert \ell - f(y)\rvert<\frac{\epsilon}{3}$.

 

[I like Serena]

We have that $x\leq c$. For each $x,y\in [a,c]$ with $|x-y|<\delta$ it follows that $\lvert f(x) - f(y)\rvert<\frac{\epsilon}{3}$. Does it hold that $|x-c|<\delta$ ?

Since we have $|x-y|<\delta$ and we are looking at the case that $x\le c \le y$, doesn't it follow that $|x-c|<\delta$? (Wondering)

 

[mathmari]

Since we have $|x-y|<\delta$ and we are looking at the case that $x\le c \le y$, doesn't it follow that $|x-c|<\delta$? (Wondering)

Ah yes. (Wasntme)

So, since $x\le c \le y$ we have that $|x-c|<|x-y|<\delta$, right? (Wondering)

Therefore we have the following:

Since $x\leq c$ we have that $|x-c|<\delta$ and so we get that $\lvert f(x) - f(c)\rvert<\frac{\epsilon}{3}$.
Since $c\geq c$ we have that $\lvert f(c) - \ell \rvert<\frac{\epsilon}{3}$.
Since $y\geq c$ we have that $\lvert \ell - f(y)\rvert<\frac{\epsilon}{3}$.

So, we get $$\lvert f(x) - f(y)\rvert \le \lvert f(x) - f(c)\rvert + \lvert f(c) - \ell \rvert + \lvert \ell - f(y)\rvert<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{3\epsilon}{3}=\epsilon$$

That means that in each case we have that $\lvert f(x) - f(y)\rvert<\epsilon$, and so $f$ is uniformly continuous, right? (Wondering)

 

[I like Serena]

That means that in each case we have that $\lvert f(x) - f(y)\rvert<\epsilon$, and so $f$ is uniformly continuous, right?

Well, technically we also have the case $y\le c\le x$, but yes, that's why $f$ is uniformly continuous. (Happy)

 

[mathmari]

Well, technically we also have the case $y\le c\le x$, but yes, that's why $f$ is uniformly continuous. (Happy)

So, we have the following:

For each $x,y\in [a,\infty)$ with $|x-y|<\delta$ we have the following cases:

• $x,y\leq c$:
  
  We have that $|f(x)-f(y)|<\frac{\epsilon}{3}<\epsilon$.
  
• $x,y\geq c$:
  
  We have that \begin{align*}|f(x)-f(y)|&=|f(x)-\ell+\ell-f(y)|=|(f(x)-\ell)-(f(y)-\ell)|\leq |f(x)-\ell|+|f(y)-\ell| \\ & <\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2\epsilon}{3}<\epsilon\end{align*}
  
• $x\leq c\leq y$:
  
  Since $x\le c \le y$ we have that $|x-c|<|x-y|<\delta$. So since $|x-c|<\delta$ it follows that $\lvert f(x) - f(c)\rvert<\frac{\epsilon}{3}$.
  
  We have the following:
  \begin{equation*}\lvert f(x) - f(y)\rvert=\lvert f(x) -f(c)+f(c)- f(y)\rvert \le \lvert f(x) -f(c)\rvert+\lvert f(c)- f(y)\rvert\end{equation*}
  
  Since $c\geq c$ it follows that $\lvert f(c) - \ell \rvert<\frac{\epsilon}{3}$.
  
  Since $y\geq c$ it follows that $\lvert \ell - f(y)\rvert<\frac{\epsilon}{3}$.
  
  So we get:
  \begin{equation*}\lvert f(x) - f(y)\rvert \le \lvert f(x) -f(c)\rvert+\lvert f(c)-\ell+\ell- f(y)\rvert\le \lvert f(x) - f(c)\rvert + \lvert f(c) - \ell \rvert + \lvert \ell - f(y)\rvert<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{3\epsilon}{3}=\epsilon\end{equation*}
  
• $y\le c\le x$ :
  
  Since $y\le c \le x$ we have that $|y-c|<|y-x|=|x-y|<\delta$. So since $|y-c|<\delta$ it follows that $\lvert f(y) - f(c)\rvert<\frac{\epsilon}{3}$.
  
  We have the following:
  \begin{equation*}\lvert f(x) - f(y)\rvert=\lvert f(y) - f(x)\rvert=\lvert f(y) -f(c)+f(c)- f(x)\rvert \le \lvert f(y) -f(c)\rvert+\lvert f(c)- f(x)\rvert\end{equation*}
  
  Since $c\geq c$ it follows that $\lvert f(c) - \ell \rvert<\frac{\epsilon}{3}$.
  
  Since $x\geq c$ it follows that $\lvert \ell - f(x)\rvert<\frac{\epsilon}{3}$.
  
  So, we get:
  \begin{equation*}\lvert f(x) - f(y)\rvert \le \lvert f(y) -f(c)\rvert+\lvert f(c)-\ell+\ell- f(x)\rvert\le \lvert f(y) - f(c)\rvert + \lvert f(c) - \ell \rvert + \lvert \ell - f(x)\rvert<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{3\epsilon}{3}=\epsilon\end{equation*}

In each case we have that $\lvert f(x) - f(y)\rvert<\epsilon$, that implies that $f$ is uniformly continuous on the interval $[a, \infty)$.
Equivalently it holds that :

Let $f: (-\infty, a] \rightarrow \mathbb{R}$ be continuous such that $\displaystyle{\lim_{x\rightarrow -\infty}f(x)=\ell}$. The function $f$ is then uniformly continuous on the interval $(-\infty, a]$.
Is everything correct? (Wondering)

 

[Euge]

Everything looks correct.

 

[mathmari]

Everything looks correct.

Thank you very much! (Happy)