# Mean value inequality?

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## Main Question or Discussion Point

The statement of the mean value inequality (MVI) is as follows:

"Let A be an open convex subset of R^n and let f:A-->R^m be continuously differentiable and such that ||Df(x)(y)||<=M||y|| for all x in A and y in R^n (i.e. the family
$(Df(x))_{x \in A}$ is uniformly lipschitz of constant M on R^n). Then for any x_1, x_2 in A, we have ||f(x_2)-f(x_1)||<=M||x_2-x_1||."

If m=1, then this is just the mean value theorem (MVT) plus the triangle inequality. But otherwise, the MVT applied to each component of f separately only leads ||f(x_2)-f(x_1)||<=mM||x_2-x_1||. So the proof suggested by the book I'm reading is that we write f(x_2)-f(x_1) using the fondamental theorem of calculus (FTC) as

$$f(x_2)-f(x_1)=\int_0^1\frac{d}{dt}f(x_1+t(x_2-x_1))dt=\int_0^1Df(x_1+t(x_2-x_1))(x_2-x_1)dt$$

and then use the triangle inequality for integrals to get the result.

But notice that the integrand is an element of R^m. So by the above, they certainly mean

$$f(x_2)-f(x_1)=\sum_{j=1}^me_j\int_0^1Df_j(x_1+t(x_2-x_1))(x_2-x_1)dt$$

which does not, to my knowledge, allows for a better conclusion than ||f(x_2)-f(x_1)||<=mM||x_2-x_1||.

Am I mistaken?

Thanks!

## Answers and Replies

fresh_42
Mentor
I have a different understanding of the mean value theorem, especially as the above doesn't mention a mean value. Anyway, it all comes down to show that
$$\left| \left|\int_0^1 Df(x_1+t(x_2-x_1))\,dt \right| \right| \leq M$$
and the left hand side is limited from above by the rectangle of $\sup Df$ and $(1-0)$, ergo $M$.