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I just had this exam question:

Let $A \subseteq \mathbb{R}$ and for $n \in \mathbb{N}$ let $f_n:A\to \mathbb{R}$. Suppose that each $f_n$ is bounded on $A$ and that $(f_n)\to f$ uniformly where $f$ is bounded. Show that $f_n$ is uniformly bounded, that is, there an $M$ such that $|f_n|\leq M$ for all $x$ in A and $n\in\mathbb{N}$.

I gave quite an elaborate answer and I now realize that this is not necessary. Is the following proof correct?

Each $f_n$ is bounded, let $M_n$ represent the bound for each $f_n$ respectively. Now define the set $$L=\{M_n: |f_n|<M_n \forall x\in A\}$$ Now let $M^*=\sup(L)$, we can do this by the axiom of completeness as $L$ is non-empty and bounded. Then $M^*\geq M_n$ for all $n\in \mathbb{N}$ and thus $M^*$ is an upper bound for each $f_n$ and all $x\in A$. Is this enough? It seems too simple. Is the assumption that $L$ is bounded correct? Again, this is not really a very deep question I was just wondering if my reasoning is sound. Thanks!

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How do you prove that $L$ is bounded? That's the key point here. And that's the original question. So you can't take it for granted. –  1015 Feb 8 '13 at 13:24
    
$f$ is bounded, let's say by $N$, now suppose that $L$ is not bounded that means there exists one $f_n$ that will exceed $N+2\epsilon$ (and beyond even) at some $x_0\in A$ for any $\epsilon >0$. This means that for this $f_n$ we will have $|f_n(x_0)-f(x_0)|>2\epsilon$ thus contradicting the fact that $(f_n)$ converges to $f$ uniformly. Would that be a correct argument to show $L$ is bounded? –  Slugger Feb 8 '13 at 13:28
    
The hypotheses are redundant. If each $f_n$ is bounded, then $f$ is. If $f$ is bounded, then each $f_n$ is. –  David Mitra Feb 8 '13 at 13:31
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If $f$ is bounded, $f_n$ doesn't have to be bounded. Just set $f_0$ as some unbounded function and add it as the first member of the sequence. –  Stefan Feb 8 '13 at 13:32
    
yeah but the question already asserts each $f_n$ to be bounded, but I need to show they are uniformly bounded. –  Slugger Feb 8 '13 at 13:34
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1 Answer

up vote 1 down vote accepted

As pointed out by David Mitra, the hypothesis that $f$ is bounded is redundant and not needed.

Taking $\epsilon=1$ in the definition of the uniform convergence of $f_n$ to $f$, we find $n_0$ such that $$ \sup_A|f_n-f|\leq 1 $$ for all $n\geq n_0$.

Now let $K$ be a bound for $|f_{n_0}|$.

Then $$ |f_n(x)|=|f_n(x)-f(x)+f(x)-f_{n_0}(x)+f_{n_0}(x)| $$ so $$ |f_n(x)|\leq |f_n(x)-f(x)|+|f(x)-f_{n_0}(x)|+|f_{n_0}(x)|\leq 1+1+K=K+2 $$ for all $x\in A$ and all $n\geq n_0$.

It only remains to take $M$ to be the maximum of $2+K$ and $\sup|f_n|$ for $n=1,\ldots,n_0-1$. This is a uniform bound for all the $|f_n|$.

Note: letting $n$ tend to $+\infty$ in the inequality above shows that $|f|$ is bounded by $K+2$.

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