From Real-Valued to Complex-Valued Extensions
The bounded Tietze extension theorem extends a continuous real-valued function from a closed subset of a normal space without increasing its range. A complex-valued function has two real coordinates, so the real theorem can be applied to each coordinate. The resulting extension is continuous, but its modulus bound must be calculated from both coordinate bounds; it does not follow by copying the original modulus bound unchanged.
Throughout, \((X,\mathcal{T})\) is a normal topological space, with the convention that normal spaces are \(T_1\) and disjoint closed sets have disjoint open neighborhoods. Let \(A\subseteq X\) be closed. The set \(\mathbb{C}\) has its usual topology, identified with \(\mathbb{R}^2\). For \(z=x+iy\), its modulus is \(|z|=\sqrt{x^2+y^2}\).
The bounded real-valued Tietze extension theorem used here says that if \(A\) is closed in a normal space and \(h:A\to\mathbb{R}\) is continuous with \(m\le h(a)\le n\) for all \(a\in A\), then \(h\) has a continuous extension \(H:X\to\mathbb{R}\) satisfying \(m\le H(x)\le n\) for all \(x\in X\). This is the earlier real-valued theorem; its hypotheses, including normality and closedness of \(A\), are essential to this application.
Proof and Bounds
Proof. First suppose \(A\ne\varnothing\). Write \(f(a)=u(a)+iv(a)\), where \(u,v:A\to\mathbb{R}\) are the real and imaginary parts. Both are continuous: they are the composites of \(f\) with the continuous coordinate projections from \(\mathbb{C}\cong\mathbb{R}^2\) to \(\mathbb{R}\). Since \(|u(a)|\le |f(a)|\le M\) and \(|v(a)|\le |f(a)|\le M\), both are bounded. Set \(M_R=\sup_{a\in A}|u(a)|\) and \(M_I=\sup_{a\in A}|v(a)|\). These are finite nonnegative real numbers, and \(-M_R\le u(a)\le M_R\), \(-M_I\le v(a)\le M_I\) for every \(a\in A\).
Apply the bounded real-valued Tietze theorem to \(u\) on the closed set \(A\), using the interval \([-M_R,M_R]\). This gives a continuous \(U:X\to\mathbb{R}\) with \(U|_A=u\) and \(|U(x)|\le M_R\) everywhere. Apply the same theorem to \(v\), using \([-M_I,M_I]\), to obtain a continuous \(V:X\to\mathbb{R}\) with \(V|_A=v\) and \(|V(x)|\le M_I\) everywhere. Define \(F(x)=U(x)+iV(x)\). The coordinate functions \(U,V\) being continuous implies that \(F\) is continuous: for any \(x_0\in X\) and \(\varepsilon>0\), continuity gives a neighborhood of \(x_0\) on which both \(|U(x)-U(x_0)|<\varepsilon/2\) and \(|V(x)-V(x_0)|<\varepsilon/2\); there, \[ |F(x)-F(x_0)|<\sqrt{(\varepsilon/2)^2+(\varepsilon/2)^2}<\varepsilon. \] On \(A\), \(F= u+iv=f\). At every \(x\in X\), \[ |F(x)|=\sqrt{U(x)^2+V(x)^2}\le\sqrt{M_R^2+M_I^2}\le\sqrt{2}\,M, \] because \(M_R\le M\) and \(M_I\le M\). If \(A=\varnothing\), choose \(F\) to be the constant zero function. It is continuous, its restriction to the empty set agrees with \(f\), and its modulus is zero, so the conclusion also holds in this case. \(\square\)
The more precise estimate uses the actual ranges of the two coordinates. If \(u(A)\subseteq[m_R,n_R]\) and \(v(A)\subseteq[m_I,n_I]\), the real theorem can instead extend within those intervals. Then \[ |F(x)|\le \sqrt{\max\{|m_R|,|n_R|\}^2+\max\{|m_I|,|n_I|\}^2}. \] This follows by bounding the absolute value of each coordinate by the larger absolute value of its interval endpoints. The estimate need not equal the largest modulus taken by \(f\) on \(A\).
Worked Example: Extending Values on Two Points
Let \(X=\mathbb{R}\) with its standard topology, \(A=\{0,1\}\), and prescribe \(f(0)=1\), \(f(1)=i\). The set \(A\) is closed, and \(f\) is continuous: every subset of \(A\) is open in its subspace topology. The input values have moduli \(|1|=1\) and \(|i|=1\), so we may take \(M=1\).
Define real-valued functions on \(\mathbb{R}\) by \[ U(t)= \begin{cases} 1,&t\le0,\\ 1-t,&0\le t\le1,\\ 0,&t\ge1, \end{cases} \qquad V(t)= \begin{cases} 0,&t\le0,\\ t,&0\le t\le1,\\ 1,&t\ge1. \end{cases} \] Each formula agrees at the shared endpoints: for \(U\), the values at \(0\) are \(1\) and \(1-0=1\), and the values at \(1\) are \(1-1=0\) and \(0\); for \(V\), they are \(0\) and \(0\) at \(0\), and \(1\) and \(1\) at \(1\). Thus both functions are continuous. Set \(F=U+iV\). Then \(F(0)=1+i0=1\) and \(F(1)=0+i1=i\).
For \(t\le0\), \(|F(t)|=|1|=1\); for \(t\ge1\), \(|F(t)|=|i|=1\). For \(0\le t\le1\), \[ |F(t)|^2=(1-t)^2+t^2=1-2t+2t^2\le1, \] since \(2t(t-1)\le0\) on that interval. Hence \(|F(t)|\le1\) for every \(t\). This example has a sharper bound than the general \(\sqrt{2}M=\sqrt{2}\).
Check: the extension takes the prescribed endpoint values, and at \(t=1/2\) its coordinates are \(U(1/2)=1/2\) and \(V(1/2)=1/2\). Its modulus is \(\sqrt{(1/2)^2+(1/2)^2}=1/\sqrt{2}\le1\), consistent with the bound.
Worked Example: Coordinate Bounds from an Interval
Let \(X=\mathbb{R}\), \(A=[-1,1]\), and \(f(x)=x+ix^2\) for \(x\in A\). The set \(A\) is closed. The function is continuous because its real and imaginary coordinates are polynomials. For \(x\in[-1,1]\), \(|x|\le1\) and \(0\le x^2\le1\), so \(|f(x)|^2=x^2+x^4\le2\). Thus \(M=\sqrt{2}\) is a valid modulus bound.
Define \(c(t)=-1\) for \(t\le-1\), \(c(t)=t\) for \(-1\le t\le1\), and \(c(t)=1\) for \(t\ge1\). The formulas agree at \(-1\) and \(1\), so \(c\) is continuous. Set \(F(t)=c(t)+ic(t)^2\). It is continuous, and if \(t\in A\), then \(c(t)=t\), giving \(F(t)=f(t)\). Everywhere, \(|c(t)|\le1\) and \(0\le c(t)^2\le1\), so \[ |F(t)|^2=c(t)^2+c(t)^4\le1+1=2. \] Therefore \(|F(t)|\le\sqrt{2}\) on all of \(\mathbb{R}\).
Check: at \(t=2\), \(c(2)=1\), so \(F(2)=1+i\) and \(|F(2)|=\sqrt{1^2+1^2}=\sqrt{2}\). The bound is attained, and on \(A\) the formula reduces exactly to \(x+ix^2\).
Worked Example: Separate Coordinate Ranges
Let \(X=\mathbb{R}\), \(A=\{0,2\}\), with \(f(0)=-2i\) and \(f(2)=1+i\). The set \(A\) is closed and finite, so \(f\) is continuous on \(A\). The two input moduli are \(|-2i|=2\) and \(|1+i|=\sqrt{2}\), so \(M=2\) is a valid bound. The real coordinates have absolute value at most \(1\), while the imaginary coordinates have absolute value at most \(2\). The coordinate estimate therefore gives \(\sqrt{1^2+2^2}=\sqrt{5}\).
For an explicit extension, define \[ U(t)= \begin{cases} 0,&t\le0,\\ t/2,&0\le t\le2,\\ 1,&t\ge2, \end{cases} \qquad V(t)= \begin{cases} -2,&t\le0,\\ -2+3t/2,&0\le t\le2,\\ 1,&t\ge2. \end{cases} \] At \(0\) the adjacent formulas give \(U=0\), \(V=-2\); at \(2\) they give \(U=1\), \(V=1\). Thus both functions are continuous. The map \(F=U+iV\) agrees with \(f\) at \(0\) and \(2\). On \([0,2]\), \(0\le U(t)\le1\) and \(-2\le V(t)\le1\); outside that interval the coordinates are the endpoint values. Consequently \(|U(t)|\le1\) and \(|V(t)|\le2\) everywhere, giving \(|F(t)|\le\sqrt{5}\).
Check: at \(t=1\), \(U(1)=1/2\) and \(V(1)=-2+3/2=-1/2\), so \(|F(1)|=\sqrt{(1/2)^2+(-1/2)^2}=1/\sqrt{2}\le\sqrt{5}\). At \(t=0\), the formula gives \(|F(0)|=|-2i|=2\), as required.
What the Bound Does—and Does Not—Say
The theorem guarantees an extension, not a unique one. Nor does the coordinate argument generally preserve the original modulus bound \(M\). From \(|f|\le M\), we know separately that \(|u|\le M\) and \(|v|\le M\). After extending these coordinates, the available estimate is \[ |F|=\sqrt{U^2+V^2}\le\sqrt{M^2+M^2}=\sqrt{2}\,M. \] Both coordinates can contribute to the modulus at once. Sharper estimates are available when the actual coordinate ranges are narrower, as in the examples.
Common pitfall. A wrong calculation is to infer \(|U+iV|\le M\) merely from \(|U|\le M\) and \(|V|\le M\). At a point where \(U=M\) and \(V=M\), the modulus is \(\sqrt{M^2+M^2}=\sqrt{2}\,M\), not \(M\) when \(M>0\). The correction is to combine the coordinate bounds with the modulus formula and state the resulting \(\sqrt{2}\,M\) estimate, or to use tighter coordinate bounds if the data provide them.
If \(M=0\), then \(f(a)=0\) for every \(a\in A\); the theorem’s coordinate bounds are both zero, and the constructed extension is identically zero. Thus the estimate also covers the zero function without a separate positive-bound assumption. If \(A\) is empty, the constant-zero extension handles the case directly.
Check Your Understanding
For each question, keep the real and imaginary coordinate bounds distinct before estimating a modulus.
- If \(|f(a)|\le3\) on a closed subset of a normal space, what uniform bound does the coordinatewise theorem guarantee for the extension?
- Suppose \(|\operatorname{Re}f|\le2\) and \(|\operatorname{Im}f|\le1/2\) on \(A\). What modulus bound follows from range-preserving real extensions?
- Why is closedness of \(A\) and normality of \(X\) relevant to this proof? Identify the real-valued theorem applied in each coordinate.
- Explain why \(|U|\le M\) and \(|V|\le M\) imply \(|U+iV|\le\sqrt{2}M\), rather than \(|U+iV|\le M\).
- What extension and bound should be used when \(A=\varnothing\), and what happens when \(M=0\)?