# What theorem is used here (Proving function is locally Lipschitz)?

Let $I, \Omega$ be intervals of the real line. Let $f: I\times \Omega \to \Bbb R$ be continuous in $I\times \Omega$, if there exists $f_x(t,x)$ in that set, and it is continuous, then $f$ is locally lipschitz on the variable $x$ in said set.

Proof:

Let $J$ be a closed bounded interval contained in $I$, let $\Omega '$ be a closed bounded interval contained in $\Omega$. Let $L= \max \{|f_x(t,x)|: t \in J, x\in \Omega '\}$. Then $$|f(t,x)-f(t,y)| = |f_x(t,\theta)(x-y)|\leq L|x-y|$$

With $\theta$ a point in the interval of extremes $x,y_\square$.

I don't get this proof, what theorem allows to the author to write that equality? I thought it was the mean value theorem, but that's for a single variable.

Could someone explain in a more detailed manner this proof?

Thanks!

• Sit on $t$ and don't let it move no matter how much it struggles. Then you have a function of one variable and the MVT can be applied as if you were still in freshman calculus. – B. S. Thomson Nov 4 '15 at 17:27
• Shouldn't I let $t$ vary? I thought that lipschitcianity (is that a word?) on $x$ would mean that that inequality holds for whatever $t$... Could you elaborate a bit? – YoTengoUnLCD Nov 4 '15 at 17:30
• After you have established this inequality by applying the calculus version of the mean value theorem to the function $x \to f(t,x)$ for each fixed value of $t$ then sit back and let $t$ run free and wild. In short: $(u,v)\to f(u,v)$ is a function of two variables but $v\to f(c,v)$ is a function of one variable (with $c$ held fixed for the moment). That was how you learned partial derivatives. Think of this as "the partial mean value theorem." You just have to "see" this point of view. – B. S. Thomson Nov 4 '15 at 21:18
• I see, let's say I have a different $L$ for each t. If I pick the maximum of all of these $L$s, that would be the one that works for all $t$ (the Lipchitz constant?). Thanks! – YoTengoUnLCD Nov 5 '15 at 0:36
• Absolutely perfect! – B. S. Thomson Nov 5 '15 at 1:03

The OP and I have cleared up the confusion here. The situation is a common one. Here we have someone quite confident on applying the mean-value theorem of the calculus in its usual form: If $g:[a,b]\to \mathbb{R}$ is continuous and $g'(x)$ exists at every interior point of the interval then, for all points $x$, $y$ in that interval and for some $\xi$ between $x$ and $y$, $$|g(x)-g(y)|=|g'(\xi)||x-y|.$$
But here it appears in a disguise: $$|f(t,x)-f(t,y)| = |f_x(t,\theta)| |x-y|$$ where the partial derivative $f_x$ is assumed to exist in a rectangle $[a,b]\times[c,d]$.