# Stone Weierstrass Theorem fails with “polynomials in sin x”

I have proven that every continuous real-valued function on $$[0,\pi]$$ is the uniform limit of a sequence of "polynomials in $$\cos x$$". That is, polynomials of the form: $$a_0+a_1\cos x+a_2\cos^2 x+a_3\cos^3 x+\cdots +a_n\cos^n x$$ by showing that all polynomials of this form, satisfy the Stone Weierstrass theorem hypothesis.

But next, I am asked to explain why why this result would fail if $$\cos(kx)$$ is replaced by $$\sin (kx)$$ which I have no idea how to show.

These are questions 26B and 26D in Bartle´s Elements of Real Analysis

• How will you get the unit element/constant functions? – cmk Oct 11 '19 at 17:38
• $\sin 0=0$ surely? – Angina Seng Oct 11 '19 at 17:39
• I would use: $a_0+a_1sin(x)$ with $a_0=0 , a_1=1$ and $x= \frac{\pi}{2}\in [0,\pi]$ – PLanderos33 Oct 11 '19 at 17:41
• That's not a constant function. You're evaluating it at $\pi/2.$ – cmk Oct 11 '19 at 17:42
• I want to note that your first statement is different that what you wrote at the end. Your first sums were in powers of $\cos x,$ whereas your later question was for $\cos kx$ and $\sin kx.$ I just looked in Bartle, and I think the meant to ask you to compare 26C and 26D – cmk Oct 11 '19 at 17:46

If you look at polynomials of the form $$\sum\limits_{k=0}^n a_k\sin (kx),$$ then you cannot get constant functions, which are required to apply Stone Weierstrass. This is because the first term is $$a_0\sin (0x)=0,$$ whereas for polynomials in $$\cos kx,$$ you had $$a_0\cos (0x)=a_0.$$

Perhaps, a more glaring problem is the failure of closure under multiplication (check trig identities). This causes the result to break down when changing from $$\cos$$ to $$\sin$$, even if you choose to allow the polynomials to include constant functions.

Even more, as pointed out by the other response, you're going to have trouble separating points (I hadn't even considered it, as I immediately saw other problems). $$\cos$$ is injective on $$[0,\pi]$$, so it has no such issues, but $$\sin$$ is not.

EDIT: I want to emphasize this, if it’s getting missed. If you don’t include constants (not clear from context if you do or not), then you’re automatically done. Even if you do allow them and consider functions of the form $$a_0+\sum\limits_{k=1}^n a_k\sin (kx),$$ the set doesn’t still form an algebra, as it’s not closed under multiplication. Also, as the other answer points out, point separation is problematic in this case.

If you want to consider things of the form $$a_0+\sum\limits_{k=1}^n a_k\sin^k(x),$$ you'll run into similar problems (the title says this, but the actual question in the body does not- the question the OP is asking about is $$26$$D in Bartle, which asks the question that I previously answered).

• One can fairly easily prove that any continuous function on $[0,\pi]$ is a uniform limit of the type $P_k(\sin x)+\cos x Q_k(\sin x), P_k, Q_k$ polynomials - follows from the result in this question for example by writing $\cos^n$ in the required way – Conrad Oct 11 '19 at 20:52

The OP asks why, if we replace $$\cos$$ with $$\sin$$ in

$$a_0+a_1\cos x+a_2\cos^2 x+a_3\cos^3 x+\cdots +a_n\cos^n x,$$

we do not get a dense subset of $$C[0,\pi].$$ The reason is not that we don't get the constant functions. Clearly we do: The functions $$a_0+0\cdot \cos x$$ capture all constant functions.

The real reason is that $$\cos x$$ is $$1–1$$ on $$[0,\pi],$$ while $$\sin x$$ is not. Note that every function $$f$$ of the form

$$\tag 1 f(x)=a_0+a_1\sin x+a_2\sin^2 x+\cdots +a_n\sin^n x$$

satisfies $$f(\pi/4)= f(3\pi/4).$$ Hence so does the uniform limit of any sequence of such functions. Since there are loads of continuous functions on $$[0,\pi]$$ that do not have this property, the functions in $$(1)$$ are not dense in $$C[0,\pi].$$

• If you look in the book, it actually asks about $\sin kx$ and $\cos kx$ (which is what the OP asks at the end). The exercise prior is for $\cos kx$ and the one before that is related to the powers of $\cos$. I suspect that Bartle referenced the wrong exercise, given the context. Not critiquing, mind. – cmk Oct 11 '19 at 20:55