If constants converge to $f$ uniformly, then f is constant.

Question

Let $f_n : [a, b] → \Bbb R$ be a sequence of constant functions (that is, for each fixed $n ∈ \Bbb N,\: f(x) = f(y)$ for all $x, y ∈ [a, b]$. Show that if $f_n → f $ uniformly, then $f$ is also a constant function.

So far what I have is that I understand the fact that a constant function means it is continuous but the opposite is not true. As such, I am trying to model my proof that when $f$ converges uniformly on an interval, then f is continuous on the same interval but I am a little unsure on how to use this foundation to prove that $f$ is constant function instead of continuous.

Answer 1

Suppose that $f(x_1)\neq f(x_2)$ and take $\varepsilon=\frac12\bigl\lvert f(x_1)-f(x_2)\bigr\rvert$. Then, if $n$ iis large enough, you have$$(\forall x\in[a,b]):\bigl\lvert f_n(x)-f(x)\bigr\rvert<\varepsilon.$$But $f_n$ is a constant function $k_n$. So, $\bigl\lvert k_n-f(x_1)\bigr\rvert<\varepsilon$ and $\bigl\lvert k_n-f(x_2)\bigr\rvert<\varepsilon$ and therefore…

Answer 2

Let $c_n = f_{n}(x)$ the sequence of constant functions. Thus, $$ |c_n - c_{m}| = |f_n(x) - f_m(x)| \leq \| f_n - f_m \|_{\infty} \rightarrow 0.$$ Hence, $(c_n)_n$ is a Cauchy sequence in $\mathbb{R}$, and there exists $c \in \mathbb{R}$ such that $\lim_n c_n = c$. Therefore, $$|f(x) - c| \leq |f(x) - f_n(x)| + |f_n(x) - c| \leq \|f_n - f \|_{\infty} + |c_n - c|.$$ So, making $n \rightarrow \infty$, we obtain that $f = c$.

Answer 3

For all $x\in [a, b]$ one has \begin{equation} f(x) = \lim_{n\to\infty} f_n(x) = \lim_{n\to\infty} f_n(a) = f(a) \end{equation}

Answer 4

As all $f_n$ are constant, nothing depends on $x$ in this question, hence $f_n$ can only be uniformly convergent to a constant $f$.

Answer 5

Fix $x_0\in [a,b]$. Then

$$ |f(x_0)-f(x)|\le |f(x_0)-f_n(x_0)|+|f_n(x_0)-f_n(x)|+|f_n(x)-f(x)| $$ Clearly $|f_n(x_0)-f_n(x)|=0$, while $|f(x_0)-f_n(x_0)|$ and $|f_n(x)-f(x)|$ can become arbitrarily small, for $n$ sufficiently large, due to the uniform convergence.