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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!

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    $\begingroup$ 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. $\endgroup$ – B. S. Thomson Nov 4 '15 at 17:27
  • $\begingroup$ 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? $\endgroup$ – YoTengoUnLCD Nov 4 '15 at 17:30
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    $\begingroup$ 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. $\endgroup$ – B. S. Thomson Nov 4 '15 at 21:18
  • $\begingroup$ 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! $\endgroup$ – YoTengoUnLCD Nov 5 '15 at 0:36
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    $\begingroup$ Absolutely perfect! $\endgroup$ – B. S. Thomson Nov 5 '15 at 1:03
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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 $[a,b]\times[c,d]$.

I think that part of the problem here (for all of us) is that when we advance from one subject to another one we tend to feel we are in a strange new world where old ideas (particularly from the elementary calculus) have to be abandoned for new methods. In analysis, in particular, I think it is best to imagine that you have a backpack filled with all the tools that you learned earlier and that all those tools can be pulled out to be used in any new situation. Sure there are some brand new ideas, but be alert and excited when something you are already an expert in gets used again.

That's how most research proceeds. We rely on techniques we have already learned and get clever as to how to twist them around to be useful in a new situation. Sometimes it's a huge twist. This one is pretty twisty.

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  • $\begingroup$ Thank you. Somehow that green tick makes it all worthwhile. $\endgroup$ – B. S. Thomson Nov 29 '15 at 0:00

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