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 such that .
- (O)Continuous functions on can be approximated uniformly by a sequence of odd polynomials (i.e., polynomials such that .

- (E) and (O) are both false
- (E) and (O) are both true
- (E) is true but (O) is false
- (E) is false but (O) is true

TIFR 2019 GS Part A, Problem 8

Weierstrass approximation theorem

Moderate

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?

\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 . Prove that it converges to on . Now think about restriction

For an odd function it is bit interesting. Consider a straight line passing through and observe that it passes through . 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}\)

\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 [See the proof of Weierstrass’ theorem]. Restrict it.

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