Uniform continuity is a stronger form of continuity. It is derived from the epsilon-delta criterion of continuity and particularly important for approximating functions.
Recap: Epsilon-delta criterion[Bearbeiten]
The definition of uniform continuity is built upon that one of the epsilon-delta criterion for continuity. We will therefore recap the epsilon-delta definition:
That is, if we choose an epsilon-tube around
, no matter how small it is, all function values in a sufficiently small environment of
will lie in it. An epsilon-tube is an interval from
to
with
around
:
A function
is continuous at
if and only if for every epsilon-tube, there is a
, such that all function values of
when evaluated between
and
fit inside the epsilon-tube:
This
may depend on the given function
, the given tube size
and from the point
. The following diagram shows an example, where a
-value is sufficiently small for
, but too large for
:
Hence, we must choose the
at
smaller. As both
-values at
and
are different, we will label them correspondingly by
and
. This shows, that the
within the definition of continuity may depend on the considered position
. The following diagram illustrates this:
Derivation of uniform continuity[Bearbeiten]
The epsilon-delta criterion guarantees that any continuous function
can be approximated. For every maximal error
and around every
we can find a
, such that the function value
at every
inside the delta-interval
around
maximally deviates from
by
. So for all of those
with
we can consider
as an approximation of
with a maximal error of
. The following diagram illustrates this approximation at several points
:
However, the
-values depend on the considered position
. This is why the rectangles in the above diagram have different sizes. For a more uniform approximation, we could impose additionally that all rectangles have to be of equal size, That means, the
-value shall be equally suitable for any
. The above diagram would then look as follows:
This is the core idea of uniform continuity. For any given
, there is a global
, such that no matter which point
is chosen, every function value
picked from the delta-interval
will be closer than
to
. This leads us to the following definition of uniform continuity for a function
, which will allow us for uniform approximations:
Definition of uniform continuity[Bearbeiten]
The formal definition of uniform continuity hence reads as follows:
In different words, for each
there is a
, such that all pairs
with
satisfy the inequality
.
Quantifier notation[Bearbeiten]
The commented version of the quantifier notation reads:
Comparing the epsilon-delta definition of uniform continuity and continuity in the quantifier notation, we notide that only two quantifiers have been permuted:
The reason is that the
used for uniform continuity is a global one, i.e. it does not depend on the position
. In order to express this independence, the existence quantifier „
“ must appear in front of the all quantifier „
“ .
Deriving the negation of uniform continuity[Bearbeiten]
The quantifier notation is a handy tool for inverting definitions - like for uniform continuity. It basically works by step-wise replacing existence and all quantifiers by the respective other expression:
So:
Negation of uniform continuity[Bearbeiten]
Uniform continuity is a global property[Bearbeiten]
Uniform continuity is a global property of a function. That means, it makes only sense to consider uniform continuity of a whole function. By contrast, continuity is a local property. We can speak of a function being continuous at a certain position
. For uniform continuity, this is not possible. However, a function can be both be globally continuous (continuous at every
) and (globally) uniformly continuous.
The globalness is just a consequence from the definition: For uniform continuity, we can find a global
, whose delta-interval around every argument is a sufficient approximation. It is no necessary to speak of a global delta at some certain argument
: we just need to find one delta for all arguments.
Recap: visaulization of the epsilon-delta criterion[Bearbeiten]
In order to visualize the epsilon-delta criterion, we will draw a rectangle centred at
with height
and width
. In order to satisfy the epsilon-delta criterion, the graph must not lie above or below the rectangle, but completely inside:
Take, for instance the square function at arguments around
. No matter how small we impose
, we can always find a
, such that the graph lies completely in the interior of the
-
-rectangle:
Conversely, if a function is not continuous, such as the sign function
, we are able to find an
, whose graph always takes values above or below the rectangle - no matter how narrow we make it. Concerning the sign function at
, this will for instance be the case if we choose
:
Visualization of uniform continuity[Bearbeiten]
For uniformly continuous functions, we can draw
one rectangle with heighth

and width

and move it to any point on the graph, without having pieces of the graph above or below the rectangle. The function

is uniformly continuous. Despite having infinite derivative at zero, the graph always runs inside the upper and lower bounds of the rectangle. However, for

this is not the case. In the vicinity of arguments near zero, the function moves infinitely far into the

-direction. No matter how narrow we choose the rectangle, the function will always break its lower and upper barriers for

close enough to zero.
For uniform continuity,
has to be the same at every
. The graph has to fit inside the rectangle, no matter where we move it. That means: For every
there must be a
, such that the
-
-rectangle can be moved arbitrarily along the graph without having function values above or below it:
For a non-uniformly continuous function, this is not possible. Anoter counter-example is the square function. In fact, for any given
(not just a specific one), we cannot find a
, such that the graph does nowhere cross the upper or lower border of the
-
-rectangle. For
-values near zero, the square function is close to constant, so the graph will fit inside some
-
-rectangle, there. But the more we move it to the right, the steeper our function will get. And at some point, the function will break the upper and lower barriers of the
-
-rectangle. So the square function in a continuous but not uniformly continuous function:
Scheme of proof: Uniform continuity[Bearbeiten]
In terms of quantifiers, the definition of uniform continuity reads:
This can be used to derive a scheme of proof for uniform continuity:
Scheme of proof: Not uniformly continuous[Bearbeiten]
In terms of quantifiers, the definition of
being not uniformly continuous reads:
This can be used to derive a scheme for disproving uniform continuity:
Uniform continuity as a stronger form of continuity[Bearbeiten]
Uniform continuity is a stronger form of continuity. That means, every uniformly continuous function is also continuous. The converse does not hold. There are functions like the square function
being continuous, but not uniformly continuous:
Every uniformly continuous function is also continuous[Bearbeiten]
The square function is continuous but not uniformly continuous[Bearbeiten]
As we have seen, every uniformly continuous function is also continuous. However, the converse does not hold. Let us consider again the square function as an example:
As we have seen in the section visualization, if there is
given, we cannot set a fixed
, such that the graph is everywhere inside the
-
-rectangle , independent from where we place it. The further we move the rectangle to the right, the steeper the square function gets. And at some point, the graph starts crossing he upper and lower borders of the
-
-rectangle.
This can also be seen by checking the epsilon-delta definition of uniform continuity:
. We want to prove the negation of this statement, i.e.
and
. Consider for instance
and assume there was some
, such that
for all real numbers
with
. Now consider some
which we will determine later. We take the centre
and some point
inside the
-interval. That means,
.
Now, we want to show that
holds.
Further,
If we choose
then it is guaranteed that
. Hence, we have proven that
is not uniformly continuous.
Example (uniformly continuous functions)
- The identity function
is uniformly continuous, since for
, we can show
by choosing
.
- Above, we have seen that the square function
is not uniformly continuous on the real numbers. However, when restricting to a compact interval, the function gets uniformly continuous. For example,
is uniformly continuous. We can prove this as follows: There is
because
. So we can choose
and then for all
with
the estimate
will hold.
- The square root function is uniformly continuous on
. Consider:
Let
be arbitrary. Then,
is a suitable choice:
Let
with
.
Without any restrictions, we may choose
.
Then, there is also
, and hence we get
.
Now, we want to prove that
.
By assumption there is
. And therefore:
So
. Since by assumption
and also
, we may take the square root of this equation and obtain
, i.e.
, which is what we wanted to show. Hence, we have proven that
is uniformly continuous.
- The following example is not uniformly continuous:
, which are sine waves oscillating faster and faster when approaching
. Assume that
was uniformly continuous. Then we could find some global
such that the epsilon-delta criterion holds everywhere. But now, for
the frequency of
gets arbitrarily high, so inside an interval close to
, there will always be a full oscillation of the sine inside any
-ball. Since the peak-to-peak amplitude of the sine function is equal to
, the condition
will never hold everywhere for
. We just have to move the rectangle close enough to 0, as illustrated by the subsequent figure:
As we have seen, not every continuous function is also uniformly continuous. But if we restrict to a closed, compact interval
, both continuity and uniform continuity will be equivalent:
Theorem (Heine-Cantor for
)
Every function
is uniformly continuous.
Proof (Heine-Cantor for
)
We choose an indirect way of proof: suppose, the function
was not uniformly continuous. That means, there is an
and for every
there are two points
, such that
but
.
The Bolzano Weierstraß theorem tells us (this is where compactness of
comes into play) that the bounded sequence
must have a convergent subsequence
, whose limit
is inside the interval
. Since
this is also the limit of the subsequence
.
Now, continuity of
implies
and
. Therefor, there must be a
, such that
and
for all
. We proceed by splitting and bounding
,which holds for all
. This contradicts our assumption
for all
. The assumption of
not being uniformly continuous must therefore have been wrong and we obtain uniform continuity.