One Approximation Step
The bounded Tietze extension theorem asks for a continuous function on an entire space that agrees with given data on a closed subset. A useful way to build such an extension is to improve an approximation in successive steps. The step proved here starts with a continuous function on a closed subset whose absolute value is bounded by \(r\). It constructs a continuous function on the whole space whose restriction leaves a smaller, uniformly bounded residual. The result is an approximation lemma, not yet a complete extension theorem.
We use the following form of Urysohn's lemma as an established result. If \((X,\mathcal{T})\) is a normal space and \(F_0,F_1\subseteq X\) are disjoint closed sets, there is a continuous function \(u:X\to[0,1]\) such that \(u=0\) on \(F_0\) and \(u=1\) on \(F_1\). Here, normal means that disjoint closed sets have disjoint open neighborhoods. The Urysohn lemma is used as a standard result from the earlier material on continuous separation; its proof is not repeated here.
The two bounds have different domains: the bound on \(g\) holds everywhere on \(X\), while the residual bound compares \(f\) and \(g\) only on \(A\). If \(r=0\), take \(g\) to be the constant zero function; then \(f=0\) on \(A\), and both bounds hold. If \(A=\varnothing\), the residual condition has no points to check, and the same choice works for any \(r\). It remains to prove the result when \(r>0\) and \(A\ne\varnothing\).
Constructing the Approximation
For \(r>0\), divide the values of \(f\) into three ranges. The low and high ranges determine two disjoint closed sets in \(X\); the values between them will be controlled by the range of the Urysohn function.
Set \(F_-=\{a\in A:f(a)\leq-r/3\}\) and \(F_+=\{a\in A:f(a)\geq r/3\}\). They are disjoint because \(r>0\), and each is closed in \(A\) by continuity of \(f\). Since \(A\) is closed in \(X\), they are closed in \(X\).
Choose continuous \(u:X\to[0,1]\) with \(u=0\) on \(F_-\) and \(u=1\) on \(F_+\).
Define \(g(x)=(2r/3)u(x)-r/3\) for \(x\in X\). This function is continuous as an affine expression in the continuous function \(u\).
Proof. The construction above gives the desired function. Since \(0\leq u(x)\leq1\), subtracting \(r/3\) after multiplying by \(2r/3\) gives $$ -\frac{r}{3}\leq g(x)\leq\frac{r}{3} $$ for every \(x\in X\), so \(|g(x)|\leq r/3\).
Now fix \(a\in A\). If \(f(a)\leq-r/3\), then \(a\in F_-\), so \(u(a)=0\) and \(g(a)=-r/3\). The assumptions \(-r\leq f(a)\leq-r/3\) imply \(-2r/3\leq f(a)+r/3\leq0\), hence \(|f(a)-g(a)|=|f(a)+r/3|\leq2r/3\). If \(f(a)\geq r/3\), then \(a\in F_+\), so \(u(a)=1\) and \(g(a)=r/3\). From \(r/3\leq f(a)\leq r\), we obtain \(0\leq f(a)-r/3\leq2r/3\), giving the same error bound.
In the remaining case, \(-r/3<f(a)<r/3\). Both \(f(a)\) and \(g(a)\) lie in the interval \([-r/3,r/3]\), whose length is \(2r/3\). Therefore their distance is at most \(2r/3\). These cases cover every \(a\in A\), including the threshold values in the first two cases. Thus \(g\) satisfies both required bounds. \(\square\)
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c,d\}\) have the discrete topology, so every subset is open and closed. Take \(A=\{a,b,c\}\), and define \(f(a)=-4\), \(f(b)=0\), and \(f(c)=5\). The function \(f\) is continuous because every subset of \(A\) is open in its subspace topology. Set \(r=6\); the values satisfy \(|f(a)|=4\leq6\), \(|f(b)|=0\leq6\), and \(|f(c)|=5\leq6\).
The threshold is \(r/3=6/3=2\). Thus \(F_-=\{a\}\), since \(-4\leq-2\), and \(F_+=\{c\}\), since \(5\geq2\). Choose \(u(a)=0\), \(u(b)=1/2\), \(u(c)=1\), and \(u(d)=1/2\). This is continuous on the discrete space, takes values in \([0,1]\), and has the required values on \(F_-\) and \(F_+\).
Here \(2r/3=2(6)/3=4\), so the formula gives \(g=4u-2\). At the four points its values are \(g(a)=4(0)-2=-2\), \(g(b)=4(1/2)-2=0\), \(g(c)=4(1)-2=2\), and \(g(d)=4(1/2)-2=0\). The global bound checks as \(|-2|=2\leq2\), \(|0|=0\leq2\), \(|2|=2\leq2\), and \(|0|=0\leq2\). On \(A\), the residuals are \(f(a)-g(a)=-4-(-2)=-2\), \(f(b)-g(b)=0-0=0\), and \(f(c)-g(c)=5-2=3\). Their absolute values \(2,0,3\) are each at most \(4=2r/3\), as required.
Check: the construction uses \(g=4u-2\), so substituting \(u=0\) gives \(-2\), and substituting \(u=1\) gives \(2\). The endpoint values agree with the two prescribed Urysohn values.
Worked Example: A Piecewise Linear Urysohn Function
Use the standard topology on \(X=\mathbb{R}\), let \(A=[-1,1]\), and set \(f(x)=3x\) on \(A\). The set \(A\) is closed, and \(f\) is continuous. For every \(x\in[-1,1]\), \(|f(x)|=3|x|\leq3\), so take \(r=3\). The threshold is \(r/3=1\). The extreme sets are \(F_-=[-1,-1/3]\) and \(F_+=[1/3,1]\), which are disjoint closed subsets of \(\mathbb{R}\).
Define \(u:\mathbb{R}\to[0,1]\) by $$ u(x)= \begin{cases} 0,&x\leq-1/3,\\ (3x+1)/2,&-1/3\leq x\leq1/3,\\ 1,&x\geq1/3. \end{cases} $$ The formulas agree at the joining points: \((3(-1/3)+1)/2=0\) and \((3(1/3)+1)/2=1\). Each piece is continuous and the values match at the joins, so \(u\) is continuous. It is zero on \(F_-\) and one on \(F_+\). Since \(2r/3=2\), the resulting function is \(g(x)=2u(x)-1\).
On \([-1,-1/3]\), \(g(x)=-1\), so the residual is \(f(x)-g(x)=3x+1\). At the ends of this interval the values are \(3(-1)+1=-2\) and \(3(-1/3)+1=0\); because this expression is increasing, it lies between \(-2\) and \(0\), and its absolute value is at most \(2\). On \([1/3,1]\), \(g(x)=1\), so the residual is \(3x-1\), increasing from \(3(1/3)-1=0\) to \(3(1)-1=2\). Its absolute value is also at most \(2\).
For \(-1/3\leq x\leq1/3\), substitution gives \(g(x)=2((3x+1)/2)-1=3x\), so \(f(x)-g(x)=3x-3x=0\). Finally \(0\leq u(x)\leq1\) gives \(-1\leq2u(x)-1\leq1\), confirming \(|g(x)|\leq1=r/3\) on all of \(\mathbb{R}\). The residual bound is \(2=2r/3\), attained at \(x=-1\) and \(x=1\).
Why the Threshold Is One Third
The choice \(r/3\) balances two possible errors. If the threshold is too small, a point in an extreme set can still differ from the prescribed value by as much as \(r-t\). If it is too large, a point between the extreme sets can differ from the chosen value by as much as \(2t\). The following calculation shows precisely how this balance works for this particular construction.
Proof. The two threshold sets are disjoint because \(t>0\). They are closed in \(X\) by continuity of \(f\) and closedness of \(A\), so Urysohn's lemma supplies \(u\). Since \(0\leq u\leq1\), the formula \(g=t(2u-1)\) gives \(-t\leq g\leq t\). On \(F_-\), \(g=-t\), and \(-r\leq f\leq-t\) gives \(|f-g|=|f+t|\leq r-t\). On \(F_+\), \(g=t\), and \(t\leq f\leq r\) gives \(|f-g|=|f-t|\leq r-t\). At any point outside these two sets, \(-t<f<t\); both \(f\) and \(g\) lie in \([-t,t]\), so \(|f-g|\leq2t\). Thus the uniform bound is \(\max\{r-t,2t\}\).
At \(t=r/3\), the two quantities are \(r-r/3=2r/3\) and \(2(r/3)=2r/3\). If \(0<t<r/3\), then \(r-t>2r/3\). If \(r/3<t\leq r\), then \(2t>2r/3\). Therefore no other allowed \(t\) gives a smaller bound in this construction, and \(t=r/3\) attains \(2r/3\). This optimality concerns the stated two-level estimate; it is not a claim that every possible approximation method must have this error. \(\square\)
Worked Example: A Threshold Choice That Does Not Give the Claimed Error
Let \(X=A=\{p\}\) have the discrete topology, let \(f(p)=-7/5\), and take \(r=3\). Then \(|f(p)|=7/5\leq3\). A tempting but incorrect claim is that thresholding at \(r/2=3/2\) and choosing a function with values in \([-r/2,r/2]\) will always give error at most \(r/2\).
For this example, \(f(p)=-7/5=-1.4\) is greater than \(-3/2=-1.5\) and less than \(3/2\). Both threshold sets are empty. The choice \(u(p)=1\) is permitted: it is continuous, takes values in \([0,1]\), and the conditions on the empty sets impose no restrictions. With the proposed range \([-3/2,3/2]\), the corresponding affine function \(g=(2(3/2))u-3/2=3u-3/2\) has \(g(p)=3(1)-3/2=3/2\). Its residual is $$ |f(p)-g(p)|=\left|-\frac{7}{5}-\frac{3}{2}\right| =\left|-\frac{14}{10}-\frac{15}{10}\right| =\frac{29}{10}. $$ Since \(29/10>3/2\), the claimed \(r/2\) error bound fails.
The one-third threshold avoids this problem. Here \(r/3=1\), and \(f(p)=-7/5\leq-1\), so the low set contains \(p\) and Urysohn's condition forces \(u(p)=0\). Then \(g(p)=2u(p)-1=-1\), and the residual is \(|-7/5-(-1)|=|-2/5|=2/5\). The theorem's bound is \(2r/3=2\), and \(2/5\leq2\). The example shows why the intermediate range and the extreme ranges must be controlled together.
How the Lemma Fits into Extension
The key output is a continuous function defined on all of \(X\), with a controlled range, while the difference from the original data is small on \(A\). The residual remains a function on \(A\); it is continuous because it is the difference of the continuous functions \(f\) and \(g|_A\). Its absolute value is bounded by \(2r/3\), rather than \(r\). This is the uniform improvement that makes the lemma useful in the construction of an extension.
A common mistake is to check only the values of \(g\) on the two extreme sets. Urysohn's conditions prescribe \(g\) there, but say nothing about its exact values between them. The proof succeeds because every intermediate \(g(a)\) is still trapped between \(-r/3\) and \(r/3\), while the intermediate value \(f(a)\) lies in the same interval. Their difference is bounded by the length \(2r/3\). Checking the whole middle range is essential to the uniform estimate.
Check Your Understanding
Use the construction and its hypotheses to answer these questions.
- Why are the two threshold sets disjoint when \(r>0\), and why does their closedness in \(A\) imply closedness in \(X\)?
- In the proof of the lemma, what value does \(g\) take on \(F_-\) and \(F_+\)?
- Why must the case \(r=0\) be handled separately from the Urysohn separation argument?
- For \(r=12\), compute the threshold, the bound on \(|g|\), and the bound on the residual.
- In the threshold proposition, explain why \(r-t\) controls the error on the extreme sets while \(2t\) controls it between them.