## 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).

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