# Proving that if $\sum_0^\infty f_n$ converges uniformly on $D$ to f and $f_n$ is bounded, then f is bounded.

"Prove that if each $f_n$ is a bounded function and $\sum_0^\infty f_n$ converges uniformly on $D$ to $f$, then $f$ is a bounded function"

I don't know how to do this at all. Any help appreciated.

• boundedness follows easily if a sequence of bounded functions $(g_n)$ converges uniformly to $f$ because 'converges uniformly' means that for any $\epsilon > 0$ there exists $N$ such that $n > N \implies$ ... – reuns Mar 30 '16 at 9:54
• Use the triangle inequality $|f| \leq |f-f_n| + |f_n|$ in conjunction with the hint above. – Winther Mar 30 '16 at 10:03

By Cauchy's Convergence Principle of uniformly convergence, $\forall\varepsilon>0\exists N>0$ s.t.$\forall n,m>N$,$|f_n-f_m|<\varepsilon,\forall x\in D$. In particular, $|f_n-f_{N+1}|<\varepsilon,\forall x\in D$. Thus $|f_n|<sup|f_{N+1}|+\varepsilon$ i.e. $f_n$ is uniformly bounded. Then you can use the definition of uniformly convergence on $f$.

Let $$\epsilon \gt 0$$ be given. $$f_n$$ converges uniformly to $$f$$ so $$\exists N \in \mathbb{N}$$ s.t.

$$|f_n - f| \lt \epsilon$$ whenever $$n \geq N$$.

For a fixed $$n = N$$,

$$|f_N - f| \lt \epsilon \implies$$ $$-\epsilon + f_N \lt f \lt \epsilon + f_N$$ --- $$(1)$$

But $$f_N$$ is bounded so $$|f_N| \leq M_N$$ (for some $$M_N \gt 0$$) $$\implies -M_N \leq f_N \leq M_N$$

So $$-\epsilon + f_N \geq -\epsilon - M_N$$ and $$\epsilon + f_N \leq \epsilon + M_N$$

So $$(1)$$ becomes $$-\epsilon - M_N \lt f \lt \epsilon + M_N$$ $$\implies |f| \lt \epsilon + M_N$$ so $$f$$ is bounded.