构造不可微Lipschitz函数及实轴上仅3点不可微的Lipschitz函数

Alright, let's break down these two Lipschitz function construction problems step by step — they're straightforward once you remember how absolute value functions behave!

The simplest and most intuitive example is the absolute value function:

f(x) = |x|

Why it's Lipschitz: #

For any real numbers (x_1) and (x_2), the triangle inequality tells us:
[||x_1| - |x_2|| \leq |x_1 - x_2|]
This means the Lipschitz constant (L = 1) satisfies the definition of a Lipschitz function ((|f(x_1) - f(x_2)| \leq L|x_1 - x_2|) for all (x_1, x_2 \in \mathbb{R})).

Why it's non-differentiable: #

At (x = 0), the left-hand derivative is:
[
\lim_{h \to 0^-} \frac{|h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1
]
The right-hand derivative is:
[
\lim_{h \to 0^+} \frac{|h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1
]
Since the left and right derivatives don't match, (f(x)) is not differentiable at (x = 0).

We can build this by adding shifted absolute value functions, which introduce non-differentiable points at their "corners". Here's the function:

f(x) = |x| + |x - 1| + |x - 2|

Why it's Lipschitz: #

Again, use the triangle inequality for each absolute value term:
[
|f(x_1) - f(x_2)| = \left| |x_1| + |x_1 - 1| + |x_1 - 2| - \left(|x_2| + |x_2 - 1| + |x_2 - 2|\right) \right|
]
[
\leq ||x_1| - |x_2|| + ||x_1 - 1| - |x_2 - 1|| + ||x_1 - 2| - |x_2 - 2||
]
[
\leq |x_1 - x_2| + |x_1 - x_2| + |x_1 - x_2| = 3|x_1 - x_2|
]
So the Lipschitz constant (L = 3) works here.

Why it's non-differentiable at exactly 3 points: #

Let's analyze the function's derivative across different intervals:

• For (x < 0): (f(x) = -x + (1 - x) + (2 - x) = 3 - 3x), so (f'(x) = -3) (differentiable)
• For (0 < x < 1): (f(x) = x + (1 - x) + (2 - x) = 3 - x), so (f'(x) = -1) (differentiable)
• For (1 < x < 2): (f(x) = x + (x - 1) + (2 - x) = x + 1), so (f'(x) = 1) (differentiable)
• For (x > 2): (f(x) = x + (x - 1) + (x - 2) = 3x - 3), so (f'(x) = 3) (differentiable)

Now check the corner points:

• At (x = 0): Left derivative = -3, Right derivative = -1 (not equal → non-differentiable)
• At (x = 1): Left derivative = -1, Right derivative = 1 (not equal → non-differentiable)
• At (x = 2): Left derivative = 1, Right derivative = 3 (not equal → non-differentiable)

All other points fall into one of the intervals above, where the derivative exists. So this function is non-differentiable at exactly 3 points ((0, 1, 2)) and Lipschitz everywhere on (\mathbb{R}).

内容的提问来源于stack exchange，提问作者rrz.math