Consider the set $\mathcal{F}= \{f:\Bbb{R} \to \Bbb{R}: |f(x)-f(y)| ≤ K|(x-y)|^a \}$ for some $a>0$ and $K>0$, then which of the following statements are true?

1. Every differentiable frunction $f : \Bbb{R} \to \Bbb{R}$ is in $\mathcal{F}$.

2. Every function $f \in \mathcal{F}$ is differentiable. (This is converse of (1))

3. Every function $f \in \mathcal{F}$ is uniformly continuous.

1. (

Since $x,y \in \mathbb{R}$ are arbitrary $|x+y|$ can also be arbirtarily large. So there cannot exist the required $K>0$ and $f \notin \mathcal{F}$.

If $f(x) = x^2$ is consider as function on $(a,b)$ a bounded interval (open, closed, semi open any kind), then the term $|x+y|$ can be bounded by $2b$ since $x,y \in (a,b)$ and we can take $k = 2b$ and $a =1$ which will make $f \in \mathcal F$.

2. (

3. (

1. Every differentiable frunction $f : \Bbb{R} \to \Bbb{R}$ is in $\mathcal{F}$.

2. Every function $f \in \mathcal{F}$ is differentiable. (This is converse of (1))

3. Every function $f \in \mathcal{F}$ is uniformly continuous.

**Solution**:1. (

**false**) Consider the function $f(x) = x^2$, then $f$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. We claim that $f \notin \mathcal{F}$. We have for any $x,y \in \mathbb{R}$, $$|x^2 - y^2| = |x-y| |x + y|$$Since $x,y \in \mathbb{R}$ are arbitrary $|x+y|$ can also be arbirtarily large. So there cannot exist the required $K>0$ and $f \notin \mathcal{F}$.

**Remark:**If $f(x) = x^2$ is consider as function on $(a,b)$ a bounded interval (open, closed, semi open any kind), then the term $|x+y|$ can be bounded by $2b$ since $x,y \in (a,b)$ and we can take $k = 2b$ and $a =1$ which will make $f \in \mathcal F$.

2. (

**false**) Consider the function $f(x) = |x|$ and this function satisfies $$||x| - |y|| \le |x - y|.$$ Hence for $k=1$ and $a=1$ this function is in $\mathcal F$. But $|x|$ is not differentiable at origin.3. (

**True**) Let $f$ be a function from $\mathcal{F}$, then $f$ satisfies $$|f(x)-f(y)| ≤ K|(x-y)|^a$$ for some $K>0$ and $a>0$. We claim that $f$ is uniformly continuous. Let $\epsilon > 0$, we find a $\delta$ which depends only on $\epsilon$ and hence the function will be uniformly continuous. We take $\delta = (\frac{\epsilon}{K})^{\frac{1}{a}}$ then $|x-y| < \delta$ implies that $|f(x) - f(y)| \leq K |x-y|^ a < K \delta^a = K ((\frac{\epsilon}{K})^{\frac{1}{a}})^a = \epsilon$. This completes the proof.**Share to your groups:****SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO SUBSCRIBE TO MY BLOG**
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