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If we divide through by $|x_1 - x_2|$, we get

$$\left|\frac{f(x_1) - f(x_2)}{x_1 - x_2}\right| \leq |x_1-x_2|,$$

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Holder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!

If we divide through by $|x_1 - x_2|$, we get

$$\left|\frac{f(x_1) - f(x_2)}{x_1 - x_2}\right| \leq |x_1-x_2|,$$

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Hölder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!

If we divide through by $|x_1 - x_2|$, we get

$|\frac{f(x_1) - f(x_2)}{x_1 - x_2}| \leq |x_1-x_2|$,

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Holder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!

If we divide through by $|x_1 - x_2|$, we get

$$\left|\frac{f(x_1) - f(x_2)}{x_1 - x_2}\right| \leq |x_1-x_2|,$$

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Holder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!

If we divide through by $|x_1 - x_2|$, we get

$|\frac{f(x_1) - f(x_2)}{x_1 - x_2}| \leq |x_1-x_2|$,

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Holder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained to me the above facts of life about Holder continuous functions with exponent $\alpha > 1$. Oops!

If we divide through by $|x_1 - x_2|$, we get

$|\frac{f(x_1) - f(x_2)}{x_1 - x_2}| \leq |x_1-x_2|$,

that is, the slope of the secant line between any two points is at most the distance between them. Fixing $x_1 = x$, taking $x_2 = x+h$ and letting $h$ approach zero shows that $f$ is differentiable at $x$ and $f'(x) = 0$. That is, $f' \equiv 0$, so by the Mean Value Theorem $f$ is constant.

The proof goes through with the right hand side of your inequality replaced by $o(|x_1-x_2|)$, so in particular if there is $\alpha > 1$ and $C > 0$ such that for all $x_1,x_2 \in \mathbb{R}$, $|f(x_1) - f(x_2)| \leq C |x_1 - x_2|^{\alpha}$. If instead we take $\alpha = 1$ we get a Lipschitz continuous function. If we take $\alpha \in (0,1)$ we get a Hölder continuous function. Such functions need not be constant, but are still very nice.

And now, an anecdote: last summer my department held a "mock AMS conference" in which all summer-supported graduate students presented short talks, the more senior of them tending to talk about their thesis work in progress. One student gave an exceptionally clear and audience-friendly talk about her work on convex subsets satisfying certain smoothness conditions on the boundary. She mentioned the prospect of proving a result for Holder continuous boundary for a certain class of exponents $\alpha \leq 1$. Casting about for a question, I decided to ask about the case of $\alpha > 1$...at which point her thesis adviser, who was sitting next to me in the audience, very politely explained the facts of life about Hölder continuous functions with exponent $\alpha > 1$. Oops!