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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 $L = \sqrt{S_1^2 + S_2^2}$.

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

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

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