Let $P_n$ be a polynomial of degree $n$. For a bounded function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$, where the infimum is taken all polynomials of degree $n$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.
I'd like to prove follwing statement:
If $Q_\lambda(x)$ is the form of $ \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$.
I would prove this by contradiction. So assume that there is no such $\Delta(Q_{\lambda_0})$, then for every $\Delta(Q_{\lambda})$, there is $\Delta(Q_{\lambda_i})$ such that $\Delta(Q_{\lambda_i})<\Delta(Q_{\lambda})$.
And I'm stuck on this point. I can't prove this without assumption that is f is continuous function. What step is needed to prove? Or Is there a constructive argument to prove? Any advice would be welcomed!