# Baby Rudin Theorem 7.26

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.

• 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. Aug 20, 2021 at 15:47
• 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 $f$). Aug 20, 2021 at 16:36
• @Mark Saving What if the following is not true $f(a)=f(b)=0$. Then how will the solution work out? Aug 21, 2021 at 13:45

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.

• 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? Aug 22, 2021 at 9:03