Continuity of Polynomial Functions

From Basic Functions to Polynomials

The identity function returns its input, and constant functions return a fixed value. Polynomials combine these simple functions by adding, multiplying, and multiplying by constants. To prove continuity for polynomials, we first need to understand how continuity behaves under those operations.

Throughout, \(E\) may be any subset of \(\mathbb{R}\). Continuity at \(a\in E\) tests only inputs \(x\in E\); the domain need not contain an interval around \(a\). The results from the previous tutorials on continuity of constant functions and the identity function provide the starting cases.

Some of the displayed coefficients may be zero. In particular, a constant function is a polynomial, and the zero function is a polynomial. The degree of a nonzero polynomial is not needed for the continuity argument: only the fact that its expression uses finitely many powers matters.

Adding and Multiplying Continuous Functions

The sum is controlled by the triangle inequality. For a product, we first keep one factor bounded near the point of continuity and then estimate the change in the other factor. This local bound is the key extra step in the product argument.

Proof. Fix \(\varepsilon>0\). Since \(f\) and \(g\) are continuous at \(a\), there are positive numbers \(\delta_f\) and \(\delta_g\) such that, for \(x\in E\), \[ |x-a|<\delta_f\ \Longrightarrow\ |f(x)-f(a)|<\frac{\varepsilon}{2}, \qquad |x-a|<\delta_g\ \Longrightarrow\ |g(x)-g(a)|<\frac{\varepsilon}{2}. \] Choose \(\delta=\min\{\delta_f,\delta_g\}\). If \(x\in E\) and \(|x-a|<\delta\), then \[ |(f+g)(x)-(f+g)(a)| \leq |f(x)-f(a)|+|g(x)-g(a)| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \] Thus \(f+g\) is continuous at \(a\).

For the product, continuity of \(f\) gives a number \(\delta_0>0\) such that \(|f(x)-f(a)|<1\) whenever \(x\in E\) and \(|x-a|<\delta_0\). Set \(M=|f(a)|+1\), so \(M\geq1\). For those inputs, \[ |f(x)|\leq |f(a)|+|f(x)-f(a)|<|f(a)|+1=M. \] By continuity of \(g\), choose \(\eta_g>0\) so that \(|g(x)-g(a)|<\varepsilon/(2M)\) when \(x\in E\) and \(|x-a|<\eta_g\). By continuity of \(f\), choose \(\eta_f>0\) so that \[ |f(x)-f(a)|<\frac{\varepsilon}{2(|g(a)|+1)} \] when \(x\in E\) and \(|x-a|<\eta_f\). Take \(\delta=\min\{\delta_0,\eta_g,\eta_f\}\). For \(x\in E\) with \(|x-a|<\delta\), the identity \[ f(x)g(x)-f(a)g(a)=f(x)\bigl(g(x)-g(a)\bigr)+g(a)\bigl(f(x)-f(a)\bigr) \] and the triangle inequality give \[ \begin{aligned} |f(x)g(x)-f(a)g(a)| &\leq |f(x)|\,|g(x)-g(a)|+|g(a)|\,|f(x)-f(a)|\\ &< M\frac{\varepsilon}{2M} +|g(a)|\frac{\varepsilon}{2(|g(a)|+1)} <\varepsilon. \end{aligned} \] This proves continuity of \(fg\) at \(a\).

Finally, if \(c=0\), then \(cf\) is constant and hence continuous. If \(c\ne0\), continuity of \(f\) supplies, for the tolerance \(\varepsilon/|c|\), a positive \(\delta\) such that \(|f(x)-f(a)|<\varepsilon/|c|\) whenever \(x\in E\) and \(|x-a|<\delta\). Then \[ |(cf)(x)-(cf)(a)|=|c|\,|f(x)-f(a)|<\varepsilon. \] Thus \(cf\) is continuous at \(a\) in either case. \(\square\)

Every Power Function Is Continuous

The constant function with value \(1\) is \(x^0\), and the identity function is \(x^1\). Each further power is obtained by multiplying the preceding power by the identity function. The product theorem therefore gives continuity of all powers.

Proof. The function \(m_0(x)=1\) is constant, so it is continuous everywhere. The function \(m_1(x)=x\) is the identity function, which is continuous everywhere by the result established in the previous tutorial. Now suppose \(k\geq1\) and \(m_k\) is continuous at every real number. Since \(m_{k+1}=m_k\,m_1\), the product theorem shows that \(m_{k+1}\) is continuous at every real number. Induction proves the claim for every nonnegative integer \(k\). \(\square\)

Proof. Each power function \(x^k\) is continuous everywhere by the preceding theorem. Multiplication by each coefficient \(c_k\) preserves continuity by the scalar-multiple part of the algebra theorem. Adding these finitely many continuous functions one at a time preserves continuity by its sum part. Therefore \(p\) is continuous at every real number. If \(a\in E\), the same epsilon-delta choice for \(p\) on \(\mathbb{R}\) works when inputs are restricted to \(x\in E\), because it works for all real \(x\) satisfying the input bound. Hence the restriction is continuous at every point of \(E\). If \(E\) is empty, there are no points at which to check continuity, so the restricted-function assertion holds vacuously. \(\square\)

Worked Estimates for Polynomial Functions

The theorem establishes continuity without requiring a separate estimate for each polynomial. Explicit estimates are still useful: they show how a delta can be chosen for a particular point and tolerance. Factoring the difference \(p(x)-p(a)\) often reveals a factor of \(x-a\); the remaining factor can then be bounded near \(a\).

Why the Algebraic Argument Matters

The continuity proof for a polynomial is modular. Once sums, products, and scalar multiples preserve continuity, a finite algebraic expression built from continuous functions is continuous. The power-function induction supplies the pieces \(x^k\), and the polynomial theorem follows by assembling them.

A common pitfall in a direct epsilon-delta proof is to treat the factor left after extracting \(x-a\) as if it were automatically bounded. For example, \(x^2-a^2=(x-a)(x+a)\), but the estimate still needs a neighborhood condition to bound \(|x+a|\). In the examples, a preliminary restriction such as \(|x-a|<1\) provided that bound. The general theorem avoids repeating such estimates for every polynomial, while the explicit calculations show how to produce a concrete delta when one is requested.

The same polynomial rule may be considered on the whole real line or restricted to a smaller set. Restriction cannot invalidate a continuity estimate that already works for all nearby real inputs; it only reduces the inputs that must be tested. Rational functions will require an additional point of care: division is defined only where the denominator is nonzero. The next tutorial addresses continuity on those domains.