In a supplement to Rudin's "Principle of Mathematical Analysis" I have stumbled upon the following exercise:
Let $m:X \rightarrow Y$ be a continuous function between metric spaces and let $\{f_n\}_{n \in Z_+}$ be a sequence of complex-valued functions defined on $Y$. Find a counter-example to the following statement: if $f_n$ converges uniformly to $f$ on $Y$ then $f_n \circ m$ converges uniformly to $f \circ m$.
My attempt
First of all, I have proved the following two facts:
- if $f_n$ converges locally uniformly to $f$ on $Y$ then $f_n \circ m$ converges locally uniformly to $f \circ m$;
- if $f_n$ converges uniformly to $f$ on $Y$ then $f_n \circ m$ converges uniformly to $f \circ m$, provided $m$ be uniformly continuous on X.
Therefore, whatever the counterxample, $m$ must be defined on a non-compact metric space and $f_n \circ m$ converges locally uniformly to $f \circ m$ on $X$.
My main idea so far is that $m$ should somehow increase the distance between consecutive functions in the sequence, however I do not know how to do it.
Edit
After trying to prove the statement it seems to me that the exercise, as it is written, is asking something non-existent. Actually, the statement can also be made stronger by taking away the continuity of $m$ (which becomes necessary only when we want to prove the analogous statement for locally uniformly convergent functions).
In fact: $$\forall n \in Z_+, \sup_X (|f_n \circ m - f \circ m|) \leq \sup_Y (|f_n - f|)$$ hence, we can conclude immediately by the M-Test.
Let me know if this proof seems right.
As always, any comment or answer is welcome and let me know if I can explain myself clearer.