What am I doing wrong in my proof?

Question

If $f$ is continuously differentiable on $[a,b]$, then show that for all $x,y\in(a,b)$ using the definition of derivative $$ |f(x)-f(y)|\leq|x-y|\sup_{x\in[a,b]}|f'(x)| $$

My attempt: Since $f'$ is continuous on $[a,b]$, it is bounded so there is $M>0$ such that $|f'(y)|\leq M$ for all $y\in[a,b]$. On the other hand, the extreme value theorem implies that $\sup_{y\in[a,b]}|f'(y)|$ exists so we can have $|f'(y)|\leq\sup_{y\in[a,b]}|f'(y)|$ for all $y\in[a,b]$. Since $f$ is differentiable at $y$ by the definition of derivative I have $$ f'(y)=\lim_{x\to y}\frac{f(x)-f(y)}{x-y} $$ By using the precise definition of limit and some manipulations I get: for $\epsilon=1$ there exists $\delta>0$ such that $$ 0<|x-y|<\delta\Longrightarrow|f(x)-f(y)|\leq(1+|f'(y)|)|x-y| $$ But because of $|f'(y)|\leq\sup_{y\in[a,b]}|f'(y)|$ for all $y\in[a,b]$, I get $$ 0<|x-y|<\delta\Longrightarrow|f(x)-f(y)|\leq\left(1+\sup_{y\in[a,b]}|f'(y)|\right) |x-y| $$ which is different from what I was supposed to prove!

Answer 1

If you keep dividing a closed interval repeatedly into two halves and pick one of the sub-intervals at each step the there will be a unique point in the intersection. (The left end points form an increasing sequence and the right end points form an decreasing sequence, so they both have a limit and the lengths of the intervals tends to 0. Hence there is a unique point in the intersection). If you are willing to use this fact then a proof can be given without any other theorem: suppose there exist $x,y$ such that $|f(x)-f(y)|>M|x-y|$ where $M>\sup \{|f'(x)|:a\leq x \leq b\}$. We may suppose $x<y$ Let $u= \frac {x+y} 2$. Then either $|f(x)-f(u)|>M|x-u|$ or $|f(u)-f(y)|>M|u-y|$. In the first case consider the interval $[x,u]$ and in the second case consider $[u,y]$. Repeating this process we get a decreasing sequence of intervals $[u_n,v_n]$ such that $|f(u_n)-f(v_n)|>M|u_n-v_n|$ for all $n$. Let $z$ be the unique point in the intersection of these intervals. Then $x_n \to z$ and $y_n \to z$. Since $\frac {f(u_n)-f(v_n)} {u_n-v_n} \to f'(z)$ we get a contradiction by letting $n \to \infty$ in the inequality $|f(u_n)-f(v_n)|>M|u_n-v_n|$

Answer 2

There is no way to prove this without using the completeness of the real numbers in some deep way. You can (prove and then) use the mean value theorem, or use a more ad hoc compactness or "real induction" type argument, but there is no getting around doing some deep analysis rather than just epsilon-pushing.

If you really don't want to first prove the mean value theorem, here's a sketch of a direct "real induction" argument you can use instead. Let me write $M=\sup_{x\in[a,b]}|f'(x)|$. Suppose WLOG that $x<y$, fix an $\epsilon>0$, and let $C$ be the set of $z\in [x,y]$ such that $|f(x)-f(z)|\leq |x-z|(M+\epsilon)$. There are a couple key observations now. First, $C$ is closed and $x\in C$. Second, if $z\in C$ and $z<y$, then there exists $\delta>0$ such that $z+\delta\in C$ (I'll leave this for you to prove, using the definition of the derivative at $z$).

Now let $z=\sup C$. Using the two properties of $C$ above, you can prove that $z=y$ (again, I leave this to you). Since $C$ is closed, this means $y\in C$, and so $|f(x)-f(y)|\leq|x-y|(M+\epsilon)$. Finally, use the fact that this holds for arbitrary $\epsilon>0$ to conclude that $|f(x)-f(y)|\leq |x-y|M$.

Answer 3

If $f$ is $C^1$ then $f(x)-f(y) = \int_x^t f'(t) dt $. Then $|f(x)-f(y)| \le \int_x^y |f'(t)|dt \le |x-y|\sup_{t \in [a,b]} |f'(t)|$.

A marginally different approach:

The key point here is that since $f$ is $C^1$ we have that $f$ is 'uniformly differentiable' in the sense that $\lim_{h \to 0} \sup_x |{f (x+h)-f(x) \over h} -f'(x)|= 0$. This follows from uniform continuity and the fact that $f(x+h)-f(x)-f'(x)h = \int_0^h (f'(t)-f'(x)) h dt$.

Let $\epsilon >0$, then there is some $\delta>0$ such that if $|h| < \delta$, then $\sup_x |{f (x+h)-f(x) \over h} -f'(x)| < \epsilon$. Hence $|f(x+h)-f(x)| \le (|f'(x)|+ \epsilon)|h| \le (\sup_t |f'(t)|+\epsilon)h$ as long as $|h| < \delta$.

Now pick $x,y$ and choose $n$ such that $h={|x-y| \over n} < \delta$, then we have \begin{eqnarray} |f(x)-f(y)| &\le& \sum_k |f(x+(k+1){y-x \over n})-f(x+k{y-x \over n})| \\ &\le& \sum_k (\sup_t |f'(t)|+\epsilon) |{x-y \over n}| \\ &=& (\sup_t |f'(t)|+\epsilon) |x-y| \end{eqnarray}