I'm curious if I've done this correctly -- please offer suggestions/corrections if not! I'm new to working in $\Bbb R^n$ so clear insights would be appreciated.

The problem:

Let $f:\Bbb R^2 \to \Bbb R$ be such that each $D_1f$ and $D_2f$ are defined everywhere and are bounded functions. Prove that $f$ is Lipschitz.

My attempt:

The definition of Lipschitz that I'm working with is that there exists some $L > 0$ such that $|f(x) -f(y)| \leq L||x-y||$.

Since $D_1f$ and $D_2f$ are bounded, there must exist:

  • $S_1 = \sup \{||D_1f(x)|| : x \in \Bbb R^2\}$
  • $S_2 = \sup \{||D_2f(x)|| : x \in \Bbb R^2\}$

Now let $a = (a_1,a_2)$ and $b = (b_1,b_2) \in \Bbb R^2$. Then we have: \begin{align*} f(a) - f(b) &= f(a_1,a_2) - f(b_1,b_2) \\ &= f(a_1,a_2) - f(a_1,b_2) + f(a_1,b_2) - f(b_1,b_2) \end{align*} Then, by the triangle inequality: $$|f(a)-f(b)| \leq |f(a_1,a_2) - (a_1,b_2)| + |f(a_1,b_2) - f(b_1,b_2)|$$ And since the partial derivatives exist everywhere in $\Bbb R^2$, we can use the one-dimensional Mean Value Theorem to show that there exists some $c$ such that: $$\frac{f(a_1,a_2)-f(a_1,b_2)}{a_2-b_2} = D_2f(a_1,c)$$ And noting how we defined $S_2$, it follows that $$|f(a_1,a_2) - f(a_1,b_2) \leq S_2 |a_2 - b_2|$$ And similarly $$|f(a_1,b_2) - f(b_1,b_2)| \leq S_1 |a_1-b_1|$$ And using the statement we got from the triangle inequality, we have that $$|f(a)-f(b)| \leq S_1|a_1-b_1| + S_2|a_2 - b_2|$$ And by the Cauchy-Schwarz inequality, we have that $$S_1|a_1-b_1| + S_2|a_2 - b_2| \leq \sqrt{S_1^2 + S_2^2}\cdot \sqrt{(a_1-b_1)^2+(a_2-b_2)^2} = \sqrt{S_1^2 + S_2^2} \cdot ||a-b||$$ Whereby $$|f(a)-f(b)| \leq \sqrt{S_1^2 + S_2^2} \cdot ||a-b||$$ So $f$ is Lipschitz with $L = \sqrt{S_1^2 + S_2^2}$.

  • $\begingroup$ In general, if we have bounded partial derivatives, then is that sufficient to show Lipschitz? $\endgroup$ Mar 5 '20 at 16:37

The proof is correct and sufficiently detailed. My personal preference is to express it in a more "modular" form by isolating an important fact:

Lemma. A function $f:\mathbb{R}^n\to \mathbb R$ is Lipschitz if and only if there exists a constant $L$ such that the restriction of $f$ to every line parallel to a coordinate axis is Lipschitz with constant $L$.

Notice that the lemma has nothing to do with partial derivatives. One direction is trivial, the other is just the triangle inequality. E.g., $$ |f(a_1,a_2)-f(b_1,b_2)| \le |f(a_1,a_2)-f(b_1,a_2)|+|f(b_1,a_2)-f(b_1,b_2)| \\ \le L|a_1-a_2|+L|b_1-b_2| \le L\sqrt{2}\|a-b\| $$ and similarly for general $n$. $\quad\Box$

Once you have the lemma, the proof of the claim in your post boils down to setting $L = \max_i(\sup |D_if|)$ and using the one-dimensional Mean Value Theorem to show that the hypothesis of the lemma is satisfied.

  • $\begingroup$ That's very helpful. Thanks for your feedback! $\endgroup$
    – Newb
    Apr 29 '15 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.