Graph of Continuous Functions : Real space to Normed linear space

Graph of Continuous Functions :

Real space to Normed linear space

M. Velrajan

We discuss how learning the closedness of the graph of a continuous real valued function on real space, by a way of asking questions, thinking mathematically and applying various concepts leads to an extension of it to metric spaces, topological spaces and normed linear spaces.    

Graph of Continuous Functions

1. Recall that to draw the graph of real functions y = f(x) first we find the values of y for

some of the values of x, for example x = 0, 1, -1, 2, -2, then we plot the points

(x, f(x)) for these values of x and then we join all these plotted points by a free hand

curve.

This is because the functions y = f(x) we consider are continuous functions.

The graph (curve) of continuous functions has no break, that is why such functions are

called continuous functions.

By the by, we note that the graph of a real function is 

the collection of all points (x, f(x)) in the xy-plane, ℝ × ℝ.

That is why, in general, 

if f is a map from a set X into a set Y then the subset 

G = {(x, f(x)) / x ∈ X} of X × Y is called the graph of f.

But what is a map from a set X into a set Y?

Yes, a map from a set X into a set Y can be defined as a relation from X into Y such

that each x ∈ X is related to a unique y ∈ Y. 

i.e. a map from a set X into a set Y is a subset f of X × Y such that

for each x ∈ X there is a unique y ∈ Y with (x, y) ∈ f. And the unique y ∈ Y with

(x, y) ∈ f is denoted by f(x).

So the map and its graph can be considered as the same.

2. The monotone, boundedness and continuity of a real valued function defined on any

subspace of the real space can be determined from their graphs.

For example, consider two functions on [0,1] with graphs given in the following figure. 

Observe that both functions are continuous and bounded but not monotone and  not

injective. And both functions assume values less than or / and greater than their values

at 0 and 1.

2.1. Hence, a real valued function f on [a,b] may assume values less than or greater than

both f(a) and f(b).

2.3. If (x, y) is not on the graph of one of the above functions, however close to the graph

may be, then we can draw a small circle with centre (x, y) which does not intersect that

graph, hence the graph of both functions is a closed subset of the plane.

In general, 

3. The graph of a continuous real valued function defined on the real space is a closed

subset of the product ℝ × ℝ.

For, 

Let G = {(x, f(x)) / x ∈ ℝ} be the graph of a continuous function f.

Let (x, y) ∈ ℝ × ℝ - G. Then y ≠ f(x). 

Let ε = |y - f(x)| / 2, U = (y - ε, y + ε) and V = (f(x) - ε, f(x) + ε).

Then U ⋂ V = Ø. 

Since f is continuous at x, there exists δ > 0 such that 

f((x - δ, x + δ)) ⊆ V. 

Let W = (x - δ, x + δ) and (a, b) ∈ W × U.

Then a ∈ W, hence f(a) ∈ f((x - δ, x + δ)) ⊆ V.  

Since b ∈ U and U ⋂ V = Ø, b ≠ f(a).

Hence (a, b) ∉ G and hence  (x, y) ∈ W × U ⊆ ℝ × ℝ - G.

Therefore, ℝ × ℝ - G is open and hence G is a closed subset of the product ℝ × ℝ.

Can we extend this to metric spaces ?

We observe that only |x - a| and (x - a, x + a) and continuity using open intervals

(x - a, x + a) are used in the proof of 3. 

As seen in an earlier post, velrajanm.blogspot.com, Real Space to Metric Space and Normed Linear Space , replacing |x - a| and (x - a, x + a) by d(x, a) and 

B(x, a) in the case of metric space, we get : 

3.1. If X and Y are metric spaces then the graph of a continuous function 

f : X → Y is a closed subset of the product X × Y.

Can we extend this to topological spaces?

Up course, we can replace (x - a, x + a) by neighbourhood of x and continuity using

open intervals (x - a, x + a) by and continuity using neighbourhoods, but in general,

for y ≠ f(x) we cannot get neighbourhoods U and V of y and f(x) such that

U ⋂ V = Ø.

Oh! This is possible if the codomain space is Hausdorff.

So,

3.2. If X and Y are topological spaces and Y is Hausdorff then the graph of a continuous

function f : X → Y is a closed subset of the product X × Y.

Write the proof by necessary replacements in the proof of 3.

 

Natural question is : 

Is Y Hausdorff necessary in 3.2. ?  

Yes.

If X is a non-Hausdorff topological space then the identity function i : X → X is

continuous, but its graph {(x, x) / x ∈ X} is not a closed subset of the product

X × X. Verify!

Next natural question is : Is the converse of 3.2. true?

We try to prove the converse of 3.2. 

Suppose X and Y are topological spaces, Y is Hausdorff and the graph

 G = {(x, f(x)) / x ∈ X} of f : X → Y is a closed subset of the product X × Y.

Think how to prove f is continuous, using G is closed in X × Y. 

So, we try to prove the continuity of f by using closed sets. 

Let A be a closed subset of Y. 

How to connect A is closed in Y with G is closed in X × Y.

Yes,

Then X × A is closed in X × Y and hence G ⋂ (X × A) is closed in X × Y.

And G ⋂ (X × A) = {(x, f(x)) / f(x) ∈ A} = {(x, f(x)) / x ∈ f - 1(A)}.

Observe the relation between f - 1(A) and G ⋂ (X × A).

f - 1(A) is the set of all first coordinates of all points in G ⋂ (X × A)

i.e. f - 1(A) = 𝞹1(G ⋂ (X × A)), where 𝞹1 : X × Y → X is the projection 

defined by 𝞹1((x, y)) = x.

The projection 𝞹1 is continuous, but f - 1(A) = 𝞹1(G ⋂ (X × A)) is closed in X does not

follow from the continuity of 𝞹1.

Think, when f - 1(A) = 𝞹1(G ⋂ (X × A)) is closed in X. 

Yes, if the projection 𝞹1 is a closed map then

f - 1(A) = 𝞹1(G ⋂ (X × A)) is closed in X and hence f is continuous.

But the projection 𝞹1 is, in general, not a closed map.

We know if Y is compact then the projection 𝞹1 : X × Y → X is a closed map.

If Y is compact then the projection 𝞹1 : X × Y → X is a closed map, 

hence f - 1(A) = 𝞹1(G ⋂ (X × A)) is closed in X and hence f is continuous.

Note that Y Hausdorff is not used to prove f is continuous.

Hence we have :

3.3. Suppose X and Y are topological spaces, Y is compact and the graph 

G = {(x, f(x)) / x ∈ X} of f : X → Y is a closed subset of the product X × Y.

Then f is continuous.

3.3.1. Is Y compact necessary in 3.3. ? 

Yes, for example,

If f : ℝ → ℝ is defined by y = f(x) = 1/x, if x ≠ 0 and f(0) = 0 then f has discontinuity

only at x = 0 but its graph is the hyperbola xy = 1 and the origin and it is a closed subset

of the product ℝ × ℝ. 

Note that ℝ is not compact, but ℝ is Hausdorff and f is not continuous.

Oh, the converse of 3.2. is not true.

By 3.2. and 3.3., we have :    

3.4. Suppose X and Y are topological spaces, Y is compact and Hausdorff and 

f : X → Y is a map. Then f is continuous if and only if

the graph G = {(x, f(x)) / x ∈ X} of f is a closed subset of the product X × Y.

Map on Algebraic structure

Now we discuss the graph of a map f from an algebraic structure to another

algebraic structure of the same nature.

Since the algebraic structures need not have topology, it is not possible to speak of

continuity of f and closedness (topological) of the graph of f.

But we can ask : Is the graph of f a substructure of the product X × Y.

 

4. Suppose X and Y are groups and f : X → Y is a map. Then

f is a homomorphism if and only if the graph of f is a subgroup of 

the product X × Y.

For,

Let G = {(x, f(x)) / x ∈ X} be the graph of f.

Suppose f is a homomorphism.

Then, for (a, f(a)), (b, f(b)) ∈ G,

(a, f(a))(b, f(b))- 1 = (a, f(a))(b- 1, f(b)- 1) 

                             = (a, f(b))(a- 1, f(b- 1))

                             = (ab- 1, f(a)f(b- 1))

                             = (ab- 1, f(ab- 1)) ∈ G.

Hence G is a subgroup of the product X × Y.

Conversely, suppose G is a subgroup of the product X × Y.

Then, for a, b ∈ X, (a, f(a)), (b, f(b)) ∈ G, hence

(a, f(a))(b, f(b)) ∈ G 

i.e. (ab, f(a)f(b)) ∈ G.

Also, since ab ∈ X,  (ab, f(ab)) ∈ G.

Since f : X → Y is a map, for each x ∈ X there is a unique y ∈ Y 

such that (x, y) ∈ G.

Hence f(ab) = f(a)f(b) and f is a homomorphism. 

5. Suppose X and Y are rings and f : X → Y is a map. 

Then f is a homomorphism if and only if the graph of f is 

a subring of the product X × Y.

6. Suppose X and Y are vector spaces over a field F and f : X → Y is a map. 

Then f is a linear transformation if and only if the graph of f is 

a subspace of the product space X × Y.

Normed Linear Spaces

7. Suppose X and Y are normed linear spaces and f : X → Y is a map. 

Then f is a linear transformation if and only if the graph of f is 

a linear subspace of the product space X × Y.

7.1. If X and Y are normed linear spaces then, by 3.1. and 7, the graph of a continuous

linear transformation f : X → Y is a closed linear subspace of the product space X × Y. 

7.2. By 3.3. and 7,

Suppose X and Y are normed linear spaces, Y is compact and 

f : X → Y is a map. If the graph G = {(x, f(x)) / x ∈ X} of f is a closed linear subspace of

the product space X × Y then f is a continuous linear transformation.

7.3. By 7.1. and 7.2,

Suppose X and Y are normed linear spaces, Y is compact and 

f : X → Y is a map. Then f is a continuous linear transformation if and only if the graph

of f is a closed linear subspace of the product space X × Y.

But in 7.3., if X and Y are Banach spaces then, by the closed graph theorem,

compactness of Y is not necessary.

7.4. Suppose X and Y are Banach spaces and f : X → Y is a map. Then f is a continuous

linear transformation if and only if the graph of f is a closed linear subspace of the

product space X × Y.