Tangent Line to Convex Graph

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Theorem

Let $f$ be a real function that is:

continuous on some closed interval $\closedint a b$
differentiable and convex on the open interval $\openint a b$.

Then all the tangent lines to $f$ are below the graph of $f$.


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Proof

Concaveup.png

Let $\TT$ be the tangent line to $f$ at some point $\tuple {c, \map f c}$, $c \in \openint a b$.

Let the gradient of $\TT$ be $m$.

Let $\tuple {x_1, y_1}$ be an arbitrary point on $\TT$.

From the point-slope form of a straight line:

\(\ds y - y_1\) \(=\) \(\ds m \paren {x - x_1}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds m \paren {x - x_1} + y_1\)


For $\TT$:

$y = \map \TT x$
$y_1 = \map f c$
$m = \map {f'} c$
$x = x$
$x_1 = c$

so:

$\map \TT x = \map {f'} c \paren {x - c} + \map f c$

Consider the graph of $f$ to the right of $\tuple {c, \map f c}$, that is, any $x$ in $\openint c b$.

Let $d$ be the directed vertical distance from $\TT$ to the graph of $f$.

That is, if $f$ is above $\TT$ then $d > 0$.

If $f$ is below $\TT$, then $d < 0$.

(From the diagram, it is apparent that $\TT$ is below $f$, but we shall prove it analytically.)

$d$ can be evaluated by:

\(\ds d\) \(=\) \(\ds \map f x - \map \TT x\)
\(\ds \) \(=\) \(\ds \map f x - \map {f'} c \paren {x - c} - \map f c\)
\(\ds \) \(=\) \(\ds \map f x - \map f c - \map {f'} c \paren {x - c}\)


By the Mean Value Theorem, there exists some constant $k$ in $\openint c b$ such that:

\(\ds \map {f'} k\) \(=\) \(\ds \frac {\map f x - \map f c} {x - c}\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} k \paren {x - c}\) \(=\) \(\ds \map f x - \map f c\)


Substitute this into the formula for $d$:

\(\ds d\) \(=\) \(\ds \map {f'} k \paren {x - c} - \map {f'} c \paren {x - c}\)
\(\ds \) \(=\) \(\ds \paren {\map {f'} k - \map {f'} c} \paren {x - c}\)


Recall that $x$ lies in the interval $\openint c b$.

So $x > c$, and the quantity $x - c$ is (strictly) positive.

$k$ is also in the interval $\openint c b$ and so $k > c$.

By construction, $f$ is convex.

By the definition of convex:

$k > c \implies \map {f'} k > \map {f'} c$

which means that:

$\paren {\map {f'} k - \map {f'} c} > 0$

Then $d$ is the product of two (strictly) positive quantities and is itself (strictly) positive.


Similarly, consider the graph of $f$ to the left of $\tuple {c, \map f c}$, that is, any $x$ in $\openint a c$.

By the same process as above, we will have:

$d = \paren {\map {f'} k - \map {f'} c} \paren {x - c}$

This time, $x < c$ and the quantity $x - c$ is (strictly) negative.

Further, $k < c$, and so by a similar argument as above:

$k < c \implies \map {f'} k < \map {f'} c$

and the quantity $\paren {\map {f'} k - \map {f'} c}$ is also (strictly) negative.

Thus $d$ will be the product of two (strictly) negative quantities, and will again be (strictly) positive.

$\blacksquare$


Also see


Sources

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