# Necessary and sufficient condition for uniform convergence.

Let $a_n$ and $b_n$ be real sequences and let $f_n(x) = a_nx + b_nx^2$ be a sequence of polynomials. What should be the necessary and sufficient conditions on the sequences $a_n$ and $b_n$ so that the $f_n$ converges uniformly to $f(x) = 0$ on $\mathbb{R}$ ?

I have some conclusions: Since the convergence is unform, $\lim_{n\rightarrow \infty}f_n(1) = \lim (a_n + b_n) = 0$. Similarly, for $x = -1$, we have $\lim(-a_n + b_n) = 0$. Adding, we get $\lim{b_n} = 0$, and so we also have $\lim a_n = 0$. But these conditions are not sufficient.

The necessary and sufficient conditions for $f_n(x) = a_n x + b_n x^2$ to converge uniformly to $f(x) \equiv 0$ is that there exists $N > 0$ such that $a_n = b_n = 0$ for $n > N$. That is, the sequences must stabilize at $0$ after finitely many terms.

To see why this is the case, assume that $f_n \rightarrow 0$ uniformly. Then in particular, we can find $N > 0$ such that if $n > N$ then

$$\sup_{x \in \mathbb{R}} |a_n x + b_n x^2| < 1.$$

However, $\sup_{x \in \mathbb{R}} |a_n x + b_n x^2| = \infty$ unless $a_n = b_n = 0$ and so we get $a_n = b_n = 0$ for all $n > N$.

• +1: By definition, the deviation about the limit $f(x)=0$ must be bounded by some read value $M$ for all $x\in \mathbb{R}$. Since the functions are polynomials, they are unbounded on $\mathbb{R}$ unless their coefficients hit exactly zero for some finite $n$. – user237392 Jun 12 '16 at 12:40
• I have 2 doubts here. Why are we saying sup of mod is less than 1. (For uniform convergence, don't we say sup should tend to 0? Are we open to have sup as 0.9?) And how did we settle on sup to be infinity? What if both lim an and lim bn are tending to 0? Can we still settle on sup to be infinite? – aarbee Nov 14 '16 at 12:43
• @Ramit: The constant $1$ is completely arbitrary. If $f_n \to 0$, then for any $\varepsilon > 0$ we must be able to find $N$ such that if $n > N$ then $\sup_{x \in \mathbb{R}} |f_n(x)| < \varepsilon$. But if $a_n \neq 0$ or $b_n \neq 0$ for some $n > N$ then $\sup_{x \in \mathbb{R}} |f_n(x)| = \infty$ which implies that in fact we must have $a_n = b_n = 0$ for all $n > N$. – levap Nov 14 '16 at 13:55
• Thanks for the clarification levap. And sorry, but one of my doubts still persists. If, at infinity, Both an and bn are tending to zero ( not exactly zero), can we still say that sup above is infinity? Because in that case, we would get 0*infinity case, which could give us any value. I wonder how we are sure that we would get sup as infinity. Thanks. – aarbee Nov 14 '16 at 14:58
• Consider an example. Let $a_n = b_n = \frac{1}{n}$ then $f_n(x) = \frac{1}{n}(x + x^2)$. For fixed $n$, we have $\lim_{x \to \infty} f_n(x) = +\infty$ (as $f_n$ is a second degree polynomial with positive coefficients) and so $\sup_{x \in \mathbb{R}} |f_n(x)| = +\infty$. – levap Nov 14 '16 at 15:13