I had to solve the following problem:
Find $0\neq f \in \mathcal{C}[-1,2]$ such that $\int_{-1}^2x^{2n}f(x) \,dx = 0$ for all $0\leq n \in \mathbb{Z}$.
Now, I have indeed managed to find such a function, and verified my answer with the solutions. But then I was thinking about the problem some more and thought to do the following and obtained a contradiction.
Take such an $f\in \mathcal{C}[-1,2]$. Note that $\lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}f(x) = 0$. By Weierstrass-Approximation Theorem, I can find a sequence of polynomials $p_k \in \mathcal{C}[-1,2], \ p_k = a_0+\cdots+a_{m_k}x^{m_k}$ such that $p_k \rightarrow f$ uniformly. Then we have, $$ \begin{align} \lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}f(x)\, dx &= \lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}\lim\limits_{k\rightarrow\infty}p_k(x)\, dx\\ &=\lim\limits_{n\rightarrow \infty}\lim\limits_{k\rightarrow\infty} \int_{-1}^2x^{2n}p_k(x) \, dx \quad \text{by uniform convergence}\\ &=\lim\limits_{k\rightarrow \infty}\lim\limits_{n\rightarrow\infty} \int_{-1}^2x^{2n}p_k(x) \, dx \quad \text{again by uniform convergence}\\ &= \lim\limits_{k\rightarrow \infty}\lim\limits_{n\rightarrow\infty} \Big(\frac{a_{m_k}}{2n+m_k+1} \ 2^{2n+m_k+1}+\cdots \Big)\\ &= \pm\infty \end{align} $$
Of course, this is a contradiction, but I can't see where I went wrong. I think it might be where I switched limits, but I am unsure.
The problem comes from the fact that you cannot perform the following switch $$ \lim_{n\to\infty}\int_{-1}^2 x^{2n}f(x)dx \not= \int_{-1}^2 \lim_{n\to\infty} x^{2n}f(x)dx$$ This is because unless $f = 0$, you will have that $x^{2n}f(x)$ will converge pointwise to a discontinuous function. Namely the function $$ g(x) = \begin{cases} f(-1) \hspace{4mm}\text{if}\hspace{4mm} x = -1\\ 0 \hspace{13mm}\text{if}\hspace{4mm} x\in (-1,1)\cup\{x:f(x)=0 \}\\ f(1) \hspace{8mm}\text{if}\hspace{4mm} x = 1\\ \infty \hspace{11mm}\text{if}\hspace{4mm} x\in (1,2]\backslash\{x:f(x)=0 \}\\ \end{cases}$$ Further, you would never be able to switch the order of integration because the Riemann integral would not exist. But you could switch the order if you know that $f\geq 0$ and if you are working with the Lebesgue integral instead by using Monotone Convergence Theorem on $[1,2]$ and Dominated Convergence Theorem on $[-1,1].$