Why do I need uniform convergence here?

Question

Yet another question about uniform convergence.

So suppose I have a sequence $(x_m)$ that converges to $x$. Let $(f_n)$ be a sequence of real valued continuous functions on a closed bounded interval $[a,b]$. Suppose $f_n$ converges to $f$.

Now I want to show $f_n(x_n)\to f(x)$.

I have proven this, but I have only used the fact $f_n$ tends to $f$ pointwise. My proof is apparently wrong as I need to use the fact that $f_n$ converges to $f$ uniformly. Why do I need it to be uniform?

$|f_n(x_n) - f(x)| \le |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)|$

So looking at the RHS, the left expression can be made arbitrarily small as $f_n$ are all continuous and $x_n \to x$, easy to prove. The right expression can also be made as small as we like since $f_n \to f$. I am already given $x$, so surely point-wise convergence is all I need to assume...

Answer 1

To give a counterexample: $x_n = 1 - 1/n$ and $f : [0,1] \to \mathbb R$ defined by $f_n(x) = x^n$. Then

$$f_n(x_n) = \left( 1 - \frac 1n\right)^n \to \frac 1e \neq 1 = f(1)$$

In your argument, we can bound $|f_n(x_n) - f_n(x)|$ independently of $f_n$ if we have uniform convergence. Otherwise (as per zhw's comment above) we only know that we can bound $|f_m(x_n) - f_m(x)|$ for each, fixed $m$.

Answer 2

Look at $f_n(x) = x^n $ on $[0,1]$ This converges pointwise to $0$ on $[0,1]$ and to $f(x) = 1$ for $x= 1$. It is now easy to find a sequence $x_n\rightarrow 1$ such that $f_n(x_n)\rightarrow 0$, which is different from $f(1)$.

Answer 3

To give a lower-tech counterexample that may explain "[why uniformity of] the convergence of $f_{n} \to f$ has to do with $|f_{n}(x_{n}) - f_{n}(x)|$ [rather than with continuity of $f_{n}$, which is given]", define $$ f_{0}(x) = \max(1 - |x - 1|, 0),\quad\text{$x$ real}, $$ and put $f_{n}(x) = f_{0}(nx)$, and $x_{n} = 1/n$. Note that

• $(x_{n}) \to 0$;

• $f_{n}(x_{n}) = f_{0}(1) = 1$ for all $n \geq 1$ ("$x_{n}$ rides the peak all the way to infinity");

• $f_{n} = 0$ off the interval $(0, 2/n)$, so $(f_{n}) \to f \equiv 0$ pointwise.

Thus $f_{n}(x_{n}) \to 1 \neq 0 = f(0)$.

The problem should be clear: Since the index in the sequence of points is the same as the index in the sequence of functions (i.e., we're considering $f_{n}(x_{n})$), there's no a priori control over "how fast" $(x_{n})$ converges compared to $(f_{n})$ without assuming $(f_{n}) \to f$ uniformly.