Let $f$ be continuous on $[a,b]$. $f$ is twice differentiable on $(a,b)$ and $|f^{\prime\prime}(x)|\leq M$ for all $x\in [a,b]$. Show $f$ is uniformly continuous.
This is a question from an exam in my university. I think this has some defect in it, because isn't any function which is continuous in a closed and bounded interval is uniformly continuous? So, I regarded $[a,b]$ as $\Bbb R$ and then I am failing. I tried to take two points $x,y$ and apply Taylor's theorem on it. And then I tried to find a constant $k$ such that $|f(x)-f(y)|\leq k|x-y|$ but haven't succeded in it yet. Can someone help me? Thanks.
Take $x_0\in (a,b).$ For any $x\in (x_0,b)$ there exists $c\in (x_0,b)$ such that
$$f'(x)-f'(x_0)=f''(c)(x-x_0).$$ Thus,
$$f'(x)=f'(x_0)+f''(c)(x-x_0)\implies |f'(x)|\le |f'(x_0)|+M|x-x_0|\le |f'(x_0)|+M(b-a).$$
In a similar way, we have that for any $x\in (a,x_0)$ it is $$|f'(x)|\le |f'(x_0)|+M(b-a).$$ (Of course, for $x=x_0$ the same bound works.)
Let's write $k=|f'(x_0)|+M(b-a).$ Now, for any $x,y\in [a,b],$ with $x<y,$ there exists $c\in (x,y)$ such that
$$f(y)-f(x)=f'(c)(y-x).$$ Since $|f'|$ is bounded by $k$ we are done.