# Continuity versus uniform continuity

We consider real-valued functions.

A real-valued function $$f : I \to \mathbb R$$ (where $$I \subseteq$$ is an interval) is continuous at $$x_0 \in I$$ when: $(\forall \epsilon > 0) (\exists \delta > 0)(\forall x \in I)(\vert x- x_0 \vert \le \delta \Rightarrow \vert f(x)- f(x_0) \vert \le \epsilon).$ When $$f$$ is continuous at all $$x \in I$$, we say that $$f$$ is continuous on $$I$$.

$$f : I \to \mathbb R$$ is said to be uniform continuity on $$I$$ if $(\forall \epsilon > 0) (\exists \delta > 0)(\forall x,y \in I)(\vert x- y \vert \le \delta \Rightarrow \vert f(x)- f(y) \vert \le \epsilon).$

Obviously, a function which is uniform continuous on $$I$$ is continuous on $$I$$. Is the converse true? The answer is negative.

### An (unbounded) continuous function which is not uniform continuous

The map $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x^2 \end{array}$ is continuous. Let’s prove that it is not uniform continuous. For $$0 < x < y$$ we have $\vert f(x)-f(y) \vert = y^2-x^2 = (y-x)(y+x) \ge 2x (y-x)$ Hence for $$y-x= \delta >0$$ and $$x = \frac{1}{\delta}$$ we get
$\vert f(x) -f(y) \vert \ge 2x (y-x) =2 > 1$ which means that the definition of uniform continuity is not fulfilled for $$\epsilon = 1$$.

For this example, the function is unbounded as $$\lim\limits_{x \to \infty} x^2 = \infty$$. Continue reading Continuity versus uniform continuity

# Around binary relations on sets

We are considering here binary relations on a set $$A$$. Let’s recall that a binary relation $$R$$ on $$A$$ is a subset of the cartesian product $$R \subseteq A \times A$$. The statement $$(x,y) \in R$$ is read as $$x$$ is $$R$$-related to $$y$$ and also denoted by $$x R y$$.

Some importants properties of a binary relation $$R$$ are:

reflexive
For all $$x \in A$$ it holds $$x R y$$
irreflexive
For all $$x \in A$$ it holds not $$x R y$$
symmetric
For all $$x,y \in A$$ it holds that if $$x R y$$ then $$y R x$$
antisymmetric
For all $$x,y \in A$$ if $$x R y$$ and $$y R x$$ then $$x=y$$
transitive
For all $$x,y,z \in A$$ it holds that if $$x R y$$ and $$y R z$$ then $$x R z$$

A relation that is reflexive, symmetric and transitive is called an equivalence relation. Let’s see that being reflexive, symmetric and transitive are independent properties.

### Symmetric and transitive but not reflexive

We provide two examples of such relations. For the first one, we take for $$A$$ the set of the real numbers $$\mathbb R$$ and the relation $R = \{(x,y) \in \mathbb R^2 \, | \, xy >0\}.$ $$R$$ is symmetric as the multiplication is also symmetric. $$R$$ is also transitive as if $$xy > 0$$ and $$yz > 0$$ you get $$xy^2 z >0$$. And as $$y^2 > 0$$, we have $$xz > 0$$ which means that $$x R z$$. However, $$R$$ is not reflexive as $$0 R 0$$ doesn’t hold.

For our second example, we take $$A= \mathbb N$$ and $$R=\{(1,1)\}$$. It is easy to verify that $$R$$ is symmetric and transitive. However $$R$$ is not reflexive as $$n R n$$ doesn’t hold for $$n \neq 1$$. Continue reading Around binary relations on sets

# A proper subspace without an orthogonal complement

We consider an inner product space $$V$$ over the field of real numbers $$\mathbb R$$. The inner product is denoted by $$\langle \cdot , \cdot \rangle$$.

When $$V$$ is a finite dimensional space, every proper subspace $$F \subset V$$ has an orthogonal complement $$F^\perp$$ with $$V = F \oplus F^\perp$$. This is no more true for infinite dimensional spaces and we present here an example.

Consider the space $$V=\mathcal C([0,1],\mathbb R)$$ of the continuous real functions defined on the segment $$[0,1]$$. The bilinear map
$\begin{array}{l|rcl} \langle \cdot , \cdot \rangle : & V \times V & \longrightarrow & \mathbb R \\ & (f,g) & \longmapsto & \langle f , g \rangle = \displaystyle \int_0^1 f(t)g(t) \, dt \end{array}$ is an inner product on $$V$$.

Let’s consider the proper subspace $$H = \{f \in V \, ; \, f(0)=0\}$$. $$H$$ is an hyperplane of $$V$$ as $$H$$ is the kernel of the linear form $$\varphi : f \mapsto f(0)$$ defined on $$V$$. $$H$$ is a proper subspace as $$\varphi$$ is not always vanishing. Let’s prove that $$H^\perp = \{0\}$$.

Take $$g \in H^\perp$$. By definition of $$H^\perp$$ we have $$\int_0^1 f(t) g(t) \, dt = 0$$ for all $$f \in H$$. In particular the function $$h : t \mapsto t g(t)$$ belongs to $$H$$. Hence
$0 = \langle h , g \rangle = \displaystyle \int_0^1 t g(t)g(t) \, dt$ The map $$t \mapsto t g^2(t)$$ is continuous, non-negative on $$[0,1]$$ and its integral on this segment vanishes. Hence $$t g^2(t)$$ is always vanishing on $$[0,1]$$, and $$g$$ is always vanishing on $$(0,1]$$. As $$g$$ is continuous, we finally get that $$g = 0$$.

$$H$$ doesn’t have an orthogonal complement.

Moreover we have
$(H^\perp)^\perp = \{0\}^\perp = V \neq H$

# A non-compact closed ball

Consider a normed vector space $$(X, \Vert \cdot \Vert)$$. If $$X$$ is finite-dimensional, then a subset $$Y \subset X$$ is compact if and only if it is closed and bounded. In particular a closed ball $$B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}$$ is always compact if $$X$$ is finite-dimensional.

### The space $$A=C([0,1],\mathbb R)$$

Consider the space $$A=C([0,1],\mathbb R)$$ of the real continuous functions defined on the interval $$[0,1]$$ endowed with the sup norm:
$\Vert f \Vert = \sup\limits_{x \in [0,1]} \vert f(x) \vert$
Is the closed unit ball $$B_1[0]$$ compact? The answer is negative and we provide two proofs.

The first one is based on open covers. For $$n \ge 1$$, we denote by $$f_n$$ the piecewise linear map defined by $\begin{cases} f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\ f_n(\frac{1}{2^n})=1 \\ f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0 \end{cases}$ All the $$f_n$$ belong to $$B_1[0]$$. Moreover for $$1 \le n < m$$ we have $$\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}$$. Hence the supports of the $$f_n$$ are disjoint and $$\Vert f_n – f_m \Vert = 1$$.

Now consider the open cover $$\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}$$. For $$x \in B_1[0]\}$$ and $$u,v \in B_{\frac{1}{2}}(x)$$, $$\Vert u -v \Vert < 1$$. Therefore, each $$B_{\frac{1}{2}}(x)$$ contains at most one $$f_n$$ and a finite subcover of $$\mathcal U$$ will contain only a finite number of $$f_n$$ proving that $$A$$ is not compact.

Second proof based on convergent subsequence. As $$A$$ is a metric space, it is enough to prove that $$A$$ is not sequentially compact. Consider the sequence of functions $$g_n : x \mapsto x^n$$. The sequence is bounded as for all $$n \in \mathbb N$$, $$\Vert g_n \Vert = 1$$. If $$(g_n)$$ would have a convergent subsequence, the subsequence would converge pointwise to the function equal to $$0$$ on $$[0,1)$$ and to $$1$$ at $$1$$. As this function is not continuous, $$(g_n)$$ cannot have a subsequence converging to a map $$g \in A$$.

### Riesz’s theorem

The non-compactness of $$A=C([0,1],\mathbb R)$$ is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space $$X$$ is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of $$A=C([0,1],\mathbb R)$$ is just standard for infinite-dimensional normed vector spaces!