Understand the problem

Consider the following two statements:
• (E)Continuous functions on $latex[1 , 2]$ can be approximated uniformly by a sequence of even polynomials (i.e., polynomials $p(x) \in \Bbb R[x]$ such that $p(-x) = p(x))$.
• (O)Continuous functions on $[1 , 2]$ can be approximated uniformly by a sequence of odd polynomials (i.e., polynomials $p(x)\in \Bbb R[x]$ such that $p(-x) = -p(x))$.
Choose the correct option below.
1. (E) and (O) are both false
2. (E) and (O) are both true
3. (E) is true but (O) is false
4. (E) is false but (O) is true
Source of the problem
TIFR 2019 GS Part A, Problem 8
Topic
Weierstrass approximation theorem
Moderate

Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle

Do you really need a hint? Try it first!

$$g(x) = \begin{cases} f(x), & \text{if x \in [1,2] } \\[3ex] f(1), & \text{if x \in [-1,1] }\\[3ex] f(-x), & \text{if x \in [-2,-1] } \end{cases}$$. This is a continuous even function. Weierstrass’s approximation theorem, there is a $$g_n \to g(x)$$. Can you form an even sequence that converges?
The even seqn of polys are $[p_n(x)+p_n(x)]/2$. Prove that it converges to $g(x)$ on $[-2,2]$. Now think about restriction
For an odd function it is bit interesting. Consider a straight line $L(x)$ passing through $f(1)$ and $-f(1)$ observe that it passes through $(0,0)$. Consider the function $$h(x) = \begin{cases} f(x), & \text{if x \in [1,2] } \\[3ex] L(x), & \text{if x \in [-1,1] }\\[3ex] -f(-x), & \text{if x \in [-2,-1] } \end{cases}$$
There exists an odd function converging to $h(x)$[See the proof of Weierstrass’ theorem]. Restrict it.

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