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12.26

Concavity up and down

Definition

Let \(f:[a,b]\rightarrow \R\), we say that \(f\) is concave up in \([a,b]\) if \(\forall x_1<x_2\in[a,b]\), the segment \(\overline{(x_1,f(x_1),(x_2,f(x_2)}\) is strictly above the graph \(\left(f_{[a,b]}\right)\) exactly in \([a,b]\)

In other words: \(\frac{f\left(x_2\right)-f(x_1)}{x_2-x_1}\left(x-x_1\right)+f\left(x_1\right),\forall x\in\left(x_1,x_2\right)>f(x)\)

Also same with \(\frac{f\left(x_2\right)-f(x_1)}{x_2-x_1}\left(x-x_2\right)+f\left(x_2\right),\forall x\in\left(x_1,x_2\right)>f(x)\)

Because for the first one \(\frac{f\left(x_2\right)-f(x_1)}{x_2-x_1}>\frac{f\left(x\right)-f\left(x_1\right)}{x-x_1}\), the second \(\frac{f\left(x_2\right)-f(x_1)}{x_2-x_1}<\frac{f\left(x_2\right)-f\left(x\right)}{x_2-x}\)

(1)image(2)image​

In particular, if \(x_1=a\) and \(x_2=a+h\), we have \(\frac{f\left(a+h\right)-f(a)}{h}\left(x-a\right)+f\left(a\right)>f(x),\forall x\in\left\lbrack a,a+h\right\rbrack\)

image​

Similarly, we define concave down: image​

Theorem

Let \(f:[\alpha,\beta]\to\mathbb{R}\) continuous and differentiable on \((\alpha,\beta)\). If \(f\) is concave up, then

  • \(\forall a\in (\alpha,\beta)\), the tangent line at \(a\) is below the graph\((f)\) in \((\alpha,\beta)\) except at \(a\)

Proof

(1)

Let \(a\in (\alpha,\beta)\), let \(a<x_1<x_2\).

Since \(f\) is concave up, then \(\frac{f\left(x_1\right)-f\left(a\right)}{x_1-a}<\frac{f\left(x_2\right)-f(a)}{x_2-a}\).

Then we define \(h(x)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\) is strictly monotonic increasing on the right of \(a\)​

Then \(f^{\prime}(a)=\lim_{x\to a^{+}}h\left(x\right)=\inf\left\lbrace h\left(x\right):x>a\right\rbrace<h\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a},\forall x>a\Rightarrow f^{\prime}\left(a\right)\left(x-a\right)+f\left(a\right)<f\left(x\right),\forall x>a\)​

Then the tangent line to the right of \(a\) is strictly lower than \(f\) for \(x>a\)​

(2)

Let \(a\in (\alpha,\beta)\), let \(x_2<x_1<a\).

Since \(f\) is concave up, then \(\frac{f\left(x_2\right)-f\left(a\right)}{x_2-a}<\frac{f\left(x_1\right)-f(a)}{x_1-a}\).

Then we define \(h(x)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\) is strictly monotonic increasing on the right of \(a\)

Then \(f^{\prime}(a)=\lim_{x\to a^{-}}h\left(x\right)=\sup\left\lbrace h\left(x\right):x<a\right\rbrace>h\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a},\forall x<a\Rightarrow f^{\prime}\left(a\right)\left(x-a\right)+f\left(a\right)<f\left(x\right),\forall x<a\)

Then the tangent line to the left of \(a\) is strictly lower than \(f\) for \(x<a\) * \(f'\) is strictly monotonic increasing on \((\alpha,\beta)\)

Let \(a<b\in (\alpha,\beta)\), N.T.P. \(f'(a)<f'(b)\)

Then since \(f^{\prime}(a)<\frac{f\left(x\right)-f\left(a\right)}{x-a}\), let \(x=b\) here, we get \(f^{\prime}(a)<\frac{f\left(b\right)-f\left(a\right)}{b-a}\)

Then since \(f^{\prime}(a)>\frac{f\left(x\right)-f\left(a\right)}{x-a}\), let \(x=a,a=b\) here, we get \(f^{\prime}(b)>\frac{f\left(b\right)-f\left(a\right)}{b-a}\)

Thus \(f'(a)<f'(b)\)​

Lemma (Relative of Rolle's)

Let \(f:[\alpha,\beta]\to\R\) continuous and differentiable on \((\alpha,\beta)\). If \(f(\alpha)=f(\beta)\) and \(f'\) is strictly monotonic increasing, then \(\alpha\) and \(\beta\) are the only global max points, that is \(f(\alpha)=f(\beta)>f(x),\forall x\in (\alpha,\beta)\)

Proof

Suppose there exists \(x_0\in (\alpha,\beta)\) such that \(f(x_0)\geq f(\alpha)=f(\beta)\)

Take \(x_M\) to be a global maximum point in \((\alpha,\beta)\), then \(f'(x_M)=0\)

Since Mean value theorem, \(\exists c\in(\alpha,x_{M}):f^{\prime}\left(c\right)=\frac{f\left(\alpha\right)-f\left(x_{M}\right)}{\alpha-x_{M}}\geq0=f^{\prime}\left(x_{M}\right)\), but \(x_1<x_M\)

Which is a contradiction to the strictly monotonic increasing.

Theorem

Let \(f:[a,b]\to \R\) continuous and differentiable on \((\alpha, \beta)\).

  • If \(f'\) is strictly monotonic increasing in \((\alpha,\beta)\), then \(f\) is concave up
  • If \(f'\) is strictly monotonic decreasing in \((\alpha,\beta)\), then \(f\) is concave down

In particular, if \(f\) is twice differentiable on \((\alpha,\beta)\), Then

  • If \(f''>0\) in \((\alpha,\beta)\), then \(f\) is concave up
  • If \(f''<0\) in \((\alpha,\beta)\), then \(f\) is concave down

Proof

Assume \(f'\) is strictly monotonic increasing, let \(a<b\in(\alpha,\beta)\)

Let \(g(x)=f(x)-\left\lbrack\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right\rbrack\), then \(g(a)=0,g(b)=0\)

Since \(g\) is continuous (since \(f\) is continuous) in \([a,b]\) and differentiable(since \(f\) is differentiable) in \((a,b)\) and \(g^{\prime}(x)=f^{\prime}(x)-\frac{f\left(b\right)-f\left(a\right)}{b-a}\) is strictly monotonic increasing

By the lemma, \(g(x)<g(a)=g(b)=0\), then \(f(x)<\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\)

Then \(\frac{f(x)-f\left(a\right)}{x-a}<\frac{f\left(b\right)-f\left(a\right)}{b-a},\forall x\in\left(a,b\right)\) and this is (1) in the definition of concave up

Similarly with concave down

Definition

Let \(f:[\alpha,\beta]\to \R\) and let \(x_0\in (\alpha,\beta)\), we say that \(x_0\) is an inflection point if \(f\) changes the concavity state at \(x_0\)

This means that \(\exists a,b\in (\alpha,\beta)\) such that \(a<x_0<b\) and \(f\) is concave up/down in \([a,x_0]\) and concave down/up in \([x_0,b]\)

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