Prove there is no bounded function $\rho\in \mathcal{R}$ on $[-1,1]$ such that $\int_{-1}^1 f(x)\rho(x) dx=f(0)$ for every continuous $f$ on $[-1,1]$

Question

Prove there is no bounded function $\rho\in \mathcal{R}$ on $[-1,1]$ such that $\int_{-1}^1 f(x)\rho(x) dx=f(0)$ for every continuous $f$ on $[-1,1]$.

My attempt so far:

First, this Riemann integral makes sense since $f$ is continuous on a closed interval, and if $f,g\in \mathcal{R}$, $fg\in \mathcal{R}$.

Suppose there exists such $\rho\in \mathcal{R}$ on $[-1,1]$. Then, by assumption, we have

$|\int_{-1}^1 f(x)\rho(x) dx-f(0)|<\epsilon$

So,

$f(0)-\epsilon<L(P,f\rho) \leq|\int_{-1}^1 f(x)\rho(x) dx-f(0)|\leq U(P,f\rho)<f(0)+\epsilon$

Consider $U(P,f\rho)<f(0)+\epsilon$. Let $P=${$x_0,...,x_{j-1},x_j,...,x_n$} where $x_{j-1}\leq 0< x_j$. WLOG, further suppose this $P$ doesn't contain any duplicates.

$U(P,f\rho)=\sum_{i=1}^n M_{fi}M_{\rho i}\Delta x_i<f(0)+\epsilon$

I have no idea how to proceed and I don't even know if I'm on the right track.

Answer 1

Hint: Let $f_n(x)=1-n|x|$ for $ |x| \leq \frac 1n$ and $0$ for $|x| >\frac 1 n$. This function is continuous and the hypothesis gives $\int_{-1/n}^{1/n}(1-n|x|)\rho(x)dx=1$. Hence, $1 \leq \int_{-1/n}^{1/n}|\rho(x)|dx$. This is true for every $n$. Now get a contradiction by showing that $\int_{-1/n}^{1/n}|\rho(x)|dx \to 0$ as $ n \to \infty$.