Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation LinksMath Help > Calculus > Calculus Theorems > Bounded Value

Bounded Value Theorem

If f is continuous on [a,b] then f is bounded on the interval.

Suppose f is not bounded on [a,b].  (That means that given any large M, there exists a number c in [a,b] such that |f(c)| > M)

For any c in [a,b], and given any positive real number, ε, there exists a positive real number, δ, such that
|f(x) - f(c)| < ε  whenever x is in [a,b] and also in (c-δ,c+δ) because f is continuous.

Since f is not bounded on [a,b], the interval can be cut in half, and f is not bounded on one of the halves.  That is, either f is not bounded on [a,a+b/2] or f is not bounded on [a+b/2,b], because if it were bounded on both halves, then it would be bounded on the whole interval.

The interval can be successively bisected, so that we obtain a sequence of successively smaller closed intervals of arbitrarily small length, converging on a single point, say, c in [a,b], on which f is unbounded.  (We know c exists, because the set of real numbers are closed -- that is, the set of reals contains the limits of all convergent sequences of real numbers.  This follows from the "completeness property" of the reals.)

Since f is continuous at c, given any positive real number, ε, there exists a positive real number, δ, such that
|f(x) - f(c)| < ε  whenever x is in [a,b] and x is also in (c-δ,c+δ).

But at least one closed sub-interval of [a,b] -- I'll call it [a',b'] -- contains c and is contained within (c-δ,c+δ), such that f is unbounded on [a',b'].

Let M = |f(c)| + ε .

Since f is unbounded on [a',b'] there exists a c' in [a',b'] such that |f(c')| > M,
and we picked M such that M = |f(c)| + ε, so |f(c')| > |f(c)| + ε .

By the triangle inequality, |f(c) - f(c')| ≥ |f(c')| - |f(c)| > ε .

But by continuity, |f(c') - f(c)| < ε, a contradiction, proving the bounded value theorem.

How is this theorem used?

The Bounded Value Theorem is used to prove the Extreme Value Theorem, that if f is continuous on [a,b] then f has a maximum and a minimum on the interval.

Internet references

Source of this proof: Cut-the-knot

Related pages in this website

Go back to Calculus Home

Triangle Inequality

Arithmetic Rules -- properties of equality, addition, multiplication, and in particular the Axioms of Real Arithmetic, including the completeness axiom.

Extreme Value Theorem --  that a continuous function on a closed interval has a maximum (and a minimum).

 

The webmaster and author of this Math Help site is Graeme McRae.