Here is Theorem 7.26 by Rudin:

If $f$ is a continuous complex function on $[a, b]$, there exists a sequence of polynomials $P_n$ such that $$ \lim_{n \to \infty} P_n(x) = f(x) $$ uniformly on $[a, b]$. If $f$ is real, the $P_n$ may be taken real.

Here is Rudin's proof:

We may assume, without loss of generality that $[a, b] = [0, 1]$. We may also assume that $f(0) = f(1) = 0$. For if the theorem is proved for this case, consider $$ g(x) = f(x) - f(0) - x [ f(1) - f(0) ] \qquad (0 \leq x \leq 1). $$ Here $g(0) = g(1) = 0$, and if $g$ can be obtained as the limit of a uniformly convergent sequence of polynomials, it is clear that the same is true for $f$, since $f-g$ is a polynomial.

Furthermore, we define $f(x)$ to be zero for $x$ outside $[0, 1]$. Then $f$ is uniformly continuous on the whole line.

We put $$\tag{47} Q_n(x) = c_n \left( 1- x^2 \right)^n \qquad (n = 1, 2, 3, \ldots), $$ where $c_n$ is chosen so that $$ \tag{48} \int_{-1}^1 Q_n(x) \ \mathrm{d} x = 1 \qquad (n = 1, 2, 3, \ldots). $$ We need some information about the order of magnitude of $c_n$. Since $$ \begin{align} \int_{-1}^1 \left( 1-x^2 \right)^n \ \mathrm{d} x = 2 \int_0^1 \left( 1-x^2 \right)^n \ \mathrm{d} x &\geq 2 \int_0^{1/\sqrt{n}} \left( 1-x^2 \right)^n \ \mathrm{d} x \\ &\geq 2 \int_0^{1/\sqrt{n}} \left( 1- n x^2 \right) \ \mathrm{d} x \\ &= \frac{4}{3 \sqrt{n} } \\ &> \frac{1}{ \sqrt{n} }, \end{align} $$ it follows from (48) that $$ \tag{49} c_n < \sqrt{n}. $$

The inequality $\left( 1-x^2 \right)^n \geq 1-nx^2$ which we used above is easily shown to be true by considering the function $$ \left( 1- x^2 \right)^n - 1+nx^2 $$ which is zero at $x= 0$ and whose derivative is positive in $(0, 1)$.

For any $\delta > 0$, (49) implies $$ \tag{50} Q_n(x) \leq \sqrt{n} \left( 1- \delta^2 \right)^n \qquad ( \delta \leq \lvert x \rvert \leq 1), $$ so that $Q_n \to 0$ uniformly in $\delta \leq \lvert x \rvert \leq 1$.

Now set $$ \tag{51} P_n(x) = \int_{-1}^1 f(x+t) Q_n (t) \ \mathrm{d} t \qquad (0 \leq x \leq 1). $$ Our assumptions about $f$ show, by a simple change of variable, that $$ P_n(x) = \int_{-x}^{1-x} f(x+t) Q_n(t) \ \mathrm{d} t = \int_0^1 f(t) Q_n(t-x) \ \mathrm{d} t, $$ and the last integral is clearly a polynomial in $x$. Thus $\left\{ P_n \right\}$ is a sequence of polynomials, which are real if $f$ is real.

Given $\varepsilon > 0$, we choose $\delta > 0$ such that $\lvert y-x \rvert < \delta$ implies $$ \lvert f(y) - f(x) \rvert < \frac{\varepsilon}{2}. $$ Let $M = \sup \lvert f(x) \rvert$. Using (48), (50), and the fact that $Q_n(x) \geq 0$, we see that for $0 \leq x \leq 1$, $$ \begin{align} \left\lvert P_n(x) - f(x) \right\rvert &= \left\lvert \int_{-1}^1 [ f(x+t) - f(x) ] Q_n(t) \ \mathrm{d} t \right\rvert \\ &\leq \int_{-1}^1 \lvert f(x+t) - f(x) \rvert Q_n(t) \ \mathrm{d} t \\ &\leq 2M \int_{-1}^{-\delta} Q_n(t) \ \mathrm{d} t + \frac{\varepsilon}{2} \int_{-\delta}^\delta Q_n(t) \ \mathrm{d} t + 2 M \int_\delta^1 Q_n(t) \ \mathrm{d} t \\ &\leq 4M \sqrt{n} \left( 1 - \delta^2 \right)^n + \frac{\varepsilon}{2} \\ &< \varepsilon \end{align} $$ for all large enough $n$, which proves the theorem.

Why do we assume that $f(0)=f(1)=0$? Also, why do we define the function $g$ and why the $f(x)$ function should be zero for $x$ outside of $[0,1]$ interval? As it's stated in the theorem $f$ is complex function which is continuous on $[a,b]$. So if we assume all these things stated above, we prove the theorem for certain continuous function and not for every case.

  • 1
    $\begingroup$ If we approximate $g(x)$ by polynomials $P_n$ then $f(x)$ can be approximated by the polynomials $P_n(x) + f(0) + x[f(1)-f(0)]$. Since $g(0) = g(1) = 0$ it makes sense to prove the theorem for these kind of functions only. $\endgroup$ Aug 20, 2021 at 15:47
  • 1
    $\begingroup$ Rudin explicitly states why. If you prove it for the case $[a, b] = [0, 1]$ with $f(0) = f(1) = 0$, then you can prove the more general result using the method Rudin described (defining $g(x) = f(x(b - a) + a) - f(a) - x (f(b) - f(a))$, which is a function with domain $[0, 1]$ with $g(0) = f(1) = 0$, getting an approximation for $g$, and using this to get an approximation for $f$). $\endgroup$ Aug 20, 2021 at 16:36
  • $\begingroup$ @Mark Saving What if the following is not true $f(a)=f(b)=0$. Then how will the solution work out? $\endgroup$
    – Lanet
    Aug 21, 2021 at 13:45

1 Answer 1


Say a function $f : [a, b] \to \mathbb{R}$, where $a < b$, "has the convergence property" if and only if there is a sequence of polynomials with real coefficients $\{P_i\}_{i \in \mathbb{N}}$ which converges uniformly to $f$ on $[a, b]$ and, furthermore, that $P_i = 0$ for all $i$ whenever $f = 0$ everywhere.

Rudin proves the following theorem:

Theorem 1: For all continuous $f : [0, 1] \to \mathbb{R}$ such that $f(0) = f(1) = 0$, $f$ has the convergence property.

It sounds like you have no questions about the proof of this theorem.

From this theorem, we can then prove the theorem

Theorem 2: For all continuous $f : [0, 1] \to \mathbb{R}$, where $a \leq b$, $f$ has the convergence property.

To prove theorem 2 from theorem 1, define $g(x) = f(x) + x(f(0) - f(1)) - f(0)$. Then $g : [0, 1] \to \mathbb{R}$ is continuous, and $g(0) = g(1) = 0$. So we can take a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ which uniformly converge to $g$ over $[0, 1]$. And furthermore, if $f$ was zero everywhere, then $g$ is zero everywhere, and hence all $P_i$ are zero everywhere.

Define $Q_i(x) = P_i(x) + f(0) + x(f(1) - f(0))$. Then we see that $\{Q_i\}_{i \in \mathbb{N}}$ converge uniformly to $g$. And if the $P_i$ are zero everywhere and $f(0) = f(1) = 0$, then the $Q_i$ are zero everywhere.

From here, we prove

Theorem 3: For all continuous $f : [a, b] \to \mathbb{R}$ where $a < b$, $f$ has the convergence property.

To prove this theorem, define $g(x) = f(x(b - a) + a)$, $g : [0, 1] \to \mathbb{R}$. Now $g$ is continuous and thus has the convergence property by Theorem 2. Take a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ which uniformly converge to $g$ over $[0, 1]$. Then define $Q_i(x) = P_i(\frac{x - a}{b - a}))$. We can show that $\{Q_i\}_{i \in \mathbb{N}}$ converges to $f$ over $[a, b]$. And if $f$ is zero everywhere, then $g$ is zero everywhere, so all $P_i$ are zero, so all $Q_i$ are zero.

Finally, we prove one last theorem:

Theorem 4: For all continuous $f : [a, b] \to \mathbb{C}$ where $a < b$, there is a sequence of polynomials $\{P_i\}_{i \in \mathbb{N}}$ with complex coefficients which uniformly converges to $f$ on $[a, b]$. If $f$ is real-valued, the $P_i$ may be taken to have real coefficients.

Proof: we can write $f(x) = g(x) + h(x) i$ where $g, h : [a, b] \to \mathbb{R}$ are both continuous. Thus, both $g$ and $h$ have the convergence property, so we can write $g$ as the uniform limit of $\{P_i\}_{i \in \mathbb{N}}$ and $h$ as the uniform limit of $\{Q_i\}_{i \in \mathbb{N}}$. Then $f$ is the uniform limit of $\{P_j + i Q_j\}_{j \in \mathbb{N}}$. Furthermore, if $f$ is real-valued, then $h$ is zero everywhere, so the $Q_j$ are zero everywhere, so $P_j + i Q_j$ is real everywhere and hence has real coefficients.

  • $\begingroup$ Great answer, thanks a lot! Is it a theorem what you have written at the very top of the answer? If so, can you tell me what is it called? $\endgroup$
    – Lanet
    Aug 22, 2021 at 9:03

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