Vrftsjtryk

Assume that $f:[a,b]tomathbb{R}_+$ is continuous, differentiable with f'(x)>0 and Lipschitz.
For $y>x$, I want to understand if
$$
f(y)-f(x) ge L (y-x)
$$
holds true where $L$ is a strictly positive constant.

The way I see it is that it is true by uniform continuity.

A similar question has been already asked, though this is more specific and with additional assumptions.

That would entail $f'(x)ge L>0$ by MVT. So for a counterexample,
choose an $f'$ on the interval that is bounded, non-negative, and has zeros.

Simple example: $f'(x)=3x^2$ on $[-1,1]$ so that $f(x)=x^3$, say.

This is actually an interesting question. It is certainly true if $f$ is $C^1$ on $[a,b]$.

Suppose that $f'(x) > 0$ for all $x in [a,b]$. By the compactness of $[a,b]$ and the continuity of $f'$, this implies $alpha stackrel{rm def}{=}inf_{x in [a,b]} f'(x) > 0 $. Since for any distinct $x<y in [a,b]$ we have that $frac{f(y) - f(x)}{y-x} = f'(lambda)$ for some $lambda in [a,b]$, it is easy to see that we also have $inf_{x <y in [a,b]} frac{f(y) - f(x)}{y-x} geq alpha > 0$. Hence for all $x<y in [a,b]$ we have that $f(y) - f(x) geq alpha(y-x)$.

However, if we drop the $C^1$ assumption, it's not true. Specifically, we can exhibit a differentiable (but not $C^1$) function $f:[0,1] to mathbb{R}$ such that the derivative $f'$ is bounded (hence $f$ is Lipschitz) and the derivative is strictly positive but nonetheless $inf_{x<y in [0,1]} frac{f(y) - f(x)}{y-x} = 0$. The function here is $$f(x) = int_{0}^{x} left(frac{sin frac{1}{t}}{1 + t^2} + 1right) dt$$ where we regard $sin frac{1}{0} = 0$ (remark: for notational simplicity, we always assume that $sin frac{1}{0} = cos frac{1}{0} = 0$ for the rest of the discussion). By the first fundamental theorem of calculus, for $x in (0,1]$ we have that $f'(x) = frac{sin frac{1}{x}}{1 + x^2} + 1$. Moreover, for $x in (0,1]$, $f'(x)$ is positive, for the simple fact that we have a lower bound $frac{-1}{1+x^2} + 1 > 0$. Since the integrand exhibits a discontinuity at $x=0$, we must verify the differentiability of $f$ at $x=0$ manually. We argue $f'(0) = 1$. Observe that $$f'(0) = lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2} + 1right) dt = 1 + lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2}right) dt $$

The tricky bit is verifying $$lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2}right) dt = 0$$

There are two steps here. First, we argue that $$lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left({sin frac{1}{t}}right) dt = 0$$

We first show that there is a globally differentiable function $g:mathbb{R} to mathbb{R}$ such that $g(0) = 0$ and $g'(x) = sin frac{1}{x}$ for all $x in mathbb{R}$. It is easy to verify that the function $g(x) = x^2cosleft(frac{1}{x}right) - int_{0}^{x} 2t cos frac{1}{t} dt$ satisfies these constraints. That $p(x) stackrel{rm def}{=} x^2cosleft(frac{1}{x}right)$ is globally differentiable with $p'(x) = sin frac{1}{x} + 2x cos frac{1}{x}$ is a standard elementary analysis exercise, and that $h(x) stackrel{rm def}{=} int_{0}^{x} 2t cos frac{1}{t} dt$ is globally differentiable with $h'(x) = 2x cos frac{1}{x}$ follows since the integrand is continuous, hence the FTC certainly applies. Hence the difference $g(x) = p(x) - h(x)$ is certainly differentiable with $g'(x) = sin frac{1}{x}$ and $g(0) = 0$. The fundamental theorem of calculus [footnote $mathbf{1}$ for specific details] implies that $g(x) = int_{0}^{x} sin frac{1}{t} dt$ for all $x in mathbb{R}$, and since $g'(0) = 0$, we have that $lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left({sin frac{1}{t}}right) dt = 0$ which is what we wanted.

The second step is this: $$left|lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2} - sin frac{1}{t}right) dtright| = lim_{epsilon to 0^{+}} left|epsilon^{-1}int_{0}^{epsilon}left(frac{-t^2 sin frac{1}{t}}{1+t^2}right) dt right|
leq lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left|frac{-t^2 sin frac{1}{t}}{1+t^2}right| dt leq lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left|frac{t^2}{1+t^2}right| dt leq lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left|frac{epsilon^2}{1+0^2}right| dt leq lim_{epsilon to 0^{+}} epsilon^{-1} epsilon^{3} = 0$$

This gives us that $$lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2} - sin frac{1}{t}right) dt = 0$$

Hence, by the linearity of the integral and the aforementioned results, we have that

$$lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2}right) dt = lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left({sin frac{1}{t}}right) dt + lim_{epsilon to 0^{+}} epsilon^{-1}int_{0}^{epsilon}left(frac{sin frac{1}{t}}{1 + t^2} - sin frac{1}{t}right) dt = 0 + 0 = 0$$

Overall, this implies that the function $f:[0,1] to mathbb{R}$ we constructed a few paragraphs ago satisfies $f'(0) = 1$ and, as mentioned earlier, $f'(x) > 0$ for $x in (0,1]$. Hence, indeed, we have that $f'(x)>0$ for all $x in [0,1]$. We claim that $inf_{x in [0,1]} f'(x) = 0$. To prove this, simply consider that the sequence $left{f'left(frac{1}{frac{1}{2}(4n-1)pi}right)right}_{n geq 2} = left{frac{-1}{1 + left(frac{1}{frac{1}{2}(4n-1)pi}right)^2} + 1right}_{n geq 2}$ is strictly decreasing and converges to $0$.

Since $inf_{x in [0,1]} f'(x) = 0$, we must also have that $inf_{x<y in [0,1]} frac{f(y) - f(x)}{y-x} = 0$, since derivatives can be approximated to arbitrary precision by difference quotients. This proves the result, and indeed implies that no positive constant $L$ exists such that $f(y) - f(x) geq L(y-x)$ for $y>x in [a,b]$.

[$mathbf{1}$]: The stronger version of the fundamental theorem of calculus (proven, for instance, in Spivak's text) tells us that if $f:mathbb{R} to mathbb{R}$ is Riemann integrable on some compact $[a,b]$ and if $F'(t) = f(t)$ for all $t in [a,b]$, then $F(b) - F(a) = int_{a}^{b} f(t) dt$. This is stronger because $f$ is not required to be continuous, and this is important here. This is why the "singularity" of $sin frac{1}{t}$ at $t=0$ is irrelevant in this particular context. Since we know $sin frac{1}{t}$ is the derivative of some other function (namely, $g$) and it's also Riemann integrable on all compact intervals $[a,b] subset mathbb{R}$, we do not require continuity to deduce that $g(x) = g(x) - g(0) = int_{0}^{x} sin frac{1}{t} dt$.