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 \[
f : & \mathbb R & \longrightarrow & \mathbb R \\
& x & \longmapsto & x^2 \end{array}\] 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:

For all \(x \in A\) it holds \(x R y\)
For all \(x \in A\) it holds not \(x R y\)
For all \(x,y \in A\) it holds that if \(x R y\) then \(y R x\)
For all \(x,y \in A\) if \(x R y\) and \(y R x\) then \(x=y\)
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
\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.

What about infinite-dimensional spaces?

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 \[
f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\
f_n(\frac{1}{2^n})=1 \\
\end{cases}\] 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!