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Mean value theorem

mean-value-theorem

One of the most fundamental theorems in mathematical analysis is the mean value theorem. Geometrically, the theorem says that somewhere between points A and B on a differentiable curve there is at least one tangent line parallel to the secant line AB.

Let’s discover together this great theorem and give it some applications. Here will use the concept of differential functions.

Statement and applications of the mean value theorem

The mean value theorem is used to prove certain regularities of differentiable functions. In fact, it is used to prove that a function is norm continuity; to demonstrate that a function is a Lipschitz function, etc.

Theorem: Let $a$ and $b$ be real numbers and le $f$ be a real-valued function continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists a real number $c\in (a,b)$ such that $f(b)-f(a)=f'(c)(b-a)$.

Some applications: Show that for any real numbers $x,y$, we have $|\sin(x)-\sin(y)|\le |x-y|$ and $|\arctan(x)-\arctan(y)|\le |x-y|$. In fact, without loss of generality, we can assume that $x<y$. The functions $t\mapsto \sin(t)$ and $t\mapsto \arctan(t)$ are continuous and differentiable on $\mathbb{R}$. Thus, according to the above theorem, there exists $c_1\in (x,y)$ and $c_2\in (x,y)$ such that $\sin(x)-\sin(y)=\cos(c_1)(x-y)$ and $\arctan(x)-\arctan(y)=\frac{1}{1+c_2^2}(x-y)$. This is because $\sin'(t)=\cos(t)$ and $\arctan'(t)=\frac{1}{1+t^2}$ for any $t\in\mathbb{R}$. Now we take the absolute and use the fact that $|\cos(c_1)|\le 1$ and $\frac{1}{1+c_2^2}\le 1$ to ends the proof.

The vectorial version of the mean value theorem

Consider a normed vector space $(E,\|\cdot\|)$. We deal with vector-valued functions of the form $f:[a,b]\to E$. Then the mean value theorem for such functions is somehow different. In fact, instead of equality, we have inequality. More precisely, we have the following result.

Theorem: Assume that a function $f:[a,b]\to E$ is continue on $[a,b]$ and differential on $(a,b)$. On the other hand, assume that there exists a constant $M>0$ such that $\|f'(t)\|\le M$ for any $t\in (a,b)$. Then $$ \|f(b)-f(a)\|\le M |a-b|.$$

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