# Prove $f$ is constant if $|f(x)-f(y)|\leq M|x-y|^\alpha$ [duplicate]

Let $\alpha > 1$ and $M \geq 0$. Suppose $f: \mathbb{R} \longrightarrow \mathbb{R}$ satisfies $|f(x)-f(y)|\leq M|x-y|^\alpha$ for all $x, y\in \mathbb{R}$.

How can we prove that $f$ is a constant function? I don't even know where to start.

## marked as duplicate by Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 13 '15 at 22:49

Fix $x\in\mathbb R$. For any $y\in\mathbb R$ distinct from $x$, we have $0\leq |\frac{f(x)-f(y)}{x-y}|\leq M|x-y|^{\alpha -1}$. Let $y$ approach $x$ and use the squeeze theorem to conclude that $f'(x)=0$.
Hint: divide by $|x-y|$. What does this tell you about the derivative of $f$?
Divide [x,y] into the n partitions and use equation of the problem then use triangle inequalities to reach this equation : $$|f(x) - f(y)| \leq \frac{M|x - y|^\alpha}{n^{\alpha-1}}$$ when $n\rightarrow \infty$ we have : $$\rightarrow f(x) = f(y)$$
Consider using the fact that $|f(x)-f(y)|\leq \sup_{x<a<y} (|(f'(a)|) (x-y)$.
• How do you know $f(x)$ is differentiable? – user223391 Sep 13 '15 at 22:45