Tanh is uniformly continous on R

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In summary, the function $f(x)=\tanh x$ is defined for all real numbers and is strictly increasing. Its limits as $x$ approaches negative and positive infinity are $-1$ and $1$, respectively. Since $f$ is continuous on $\mathbb{R}$, its range is $(-1,1)$. To show that $f$ is uniformly continuous on $\mathbb{R}$, we can choose a suitable $\delta$ and use the identity $f(x)-f(y)=[1-f(x)f(y)]f(x-y)$, along with the fact that $f$ is bounded by $1$. Another way to show uniform continuity is by noting that $f$ is continuous on
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
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 $1$.
 
  • #3
Euge said:
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)
 
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  • #4
mathmari said:
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 $\Bbb R$.
 
  • #5
Euge said:
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)
 
  • #6
mathmari said:
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)
 
  • #7
I like Serena said:
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)
 
  • #8
mathmari said:
What could we do in this case? (Wondering)

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

mathmari said:
Are the first two cases correct? (Wondering)

Yes, they're correct.
 
  • #9
Euge said:
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 $\lvert \ell - f(y)\rvert<\frac{\epsilon}{3}$.
 
  • #10
mathmari said:
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)
 
  • #11
I like Serena said:
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)
 
  • #12
mathmari said:
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)
 
  • #13
I like Serena said:
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)
 
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  • #14
Everything looks correct.
 
  • #15
Euge said:
Everything looks correct.

Thank you very much! (Happy)
 

Related to Tanh is uniformly continous on R

What does it mean for a function to be uniformly continuous?

Uniform continuity is a property of a function that means the function's rate of change remains consistent throughout its domain. In other words, the function does not have any sudden or abrupt changes in its behavior.

How is uniform continuity different from regular continuity?

Uniform continuity is a stronger condition than regular continuity. While both properties require a function to be continuous, uniform continuity requires that the rate of change of the function is consistent throughout its domain, whereas regular continuity only requires the function to be continuous at each point.

What is the domain of the tanh function?

The tanh function, also known as the hyperbolic tangent function, has a domain of all real numbers, or R. This means that it can take any real number as an input and produce a corresponding output.

What is the range of the tanh function?

The range of the tanh function is also all real numbers, or R. This means that the function can produce any real number as an output.

How does the uniform continuity of tanh on R impact its behavior?

The uniform continuity of tanh on R means that the function does not have any sudden or abrupt changes in its behavior. This allows us to make predictions and analyze the function's behavior with more confidence, as we can expect the same rate of change throughout its domain.

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