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Let $f_n \rightarrow f$ and $g_n \rightarrow g$ in $C([0,1])$. Is it true that $f_n \circ g_n \rightarrow f \circ g$?

where the metric space we're interested in is $(C([0,1]), d)$ and $d$ is the sup norm:

$$ d(f,g) = \sup_{x \in [0,1]} \big|f(x) - g(x)\big|$$

I've written the following proof that seems correct to me but I don't quite feel right because I didn't have to use continuity at all:

Fix $\varepsilon > 0$. Since $g^{-1}_n([0, 1]) \subset [0,1]$ and $g^{-1}([0,1]) \subset [0,1]$,

$$\sup_{x \in [0,1]} \big|f_n(g_n(x)) - f(g(x))\big| \le \sup_{x \in [0,1]} \big|f_n(x) - f(x)\big| < \varepsilon$$

for all $n \ge N$ for some $N$ since $f_n \rightarrow f$. So since $\varepsilon$ was arbitrarily small, it must be the case that $f_n \circ g_n \rightarrow f \circ g$.

Is this ok, or is my intuition that this was too easy correct and my proof is wrong?

Edit: it wasn't specified that $g_n: [0,1] \rightarrow [0,1]$, but since we are composing it with a function defined on the unit interval we can assume that the preimage of each $g_n$ is a subset of $[0,1]$.

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fxxer
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  • Are you composing the functions or multiplying them? If we're composing them: are we given that $g_n:[0,1]\to[0,1]$? Also, to make the "$\circ$" instead of the "$\cdot$", use \circ rather than \cdot. – Ben Grossmann Nov 19 '13 at 15:54

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Is this ok, or is my intuition that this was too easy correct and my proof is wrong?

Your intuition is correct. You don't have

$$\sup_{x \in [0,1]} \big|f_n(g_n(x)) - f(g(x))\big| \le \sup_{x \in [0,1]} \big|f_n(x) - f(x)\big| < \varepsilon$$

because you're evaluating $f_n$ and $f$ at different points.

You must transform $\left\lvert f_n(g_n(x)) - f(g(x))\right\rvert$ in a way that you get expressions that you can handle directly with what you have got. You can handle $f_n(y) - f(y)$ for any $y\in [0,1]$, so let's create something of that form by adding a suitable $0$:

$$\begin{align} \left\lvert f_n(g_n(x)) - f(g(x))\right\rvert &= \left\lvert f_n(g_n(x)) - f(g_n(x)) + f(g_n(x)) - f(g(x))\right\rvert\\ &\leqslant \left\lvert f_n(g_n(x)) - f(g_n(x))\right\rvert + \left\lvert f(g_n(x)) - f(g(x))\right\rvert \end{align}$$

Now you can handle the first term with $f_n \to f$, and for the second term, you use the uniform continuity of $f$ and the convergence $g_n\to g$.

Daniel Fischer
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