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Prove that if $(f_k)$ is a uniformly convergent sequence of continuous real-valued functions on a compact domain $D\subseteq \mathbb{R}$, then there is some $M\geq 0$ such that $\left|f_k(x)\right|\leq M$ for every $x\in D$ and for every $k\in \mathbb{N}$.

My response: Basically, I am trying to show that uniform convergence on a compact domain implies uniform boundedness. Let $f(x)$ be the limiting function. Then I know that $\lim_{k\to\infty} \sup_{x \in D} | f_k (x) - f(x) | = 0$. Also, I know that $f$ is continuous, therefore it attains an absolute maximum $\in D$. How can I apply these two things to prove it?

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    Since the sequence $\|f_k-f\|_\infty$ is convergent, it is bounded : $\|f_k-f\|_\infty\leq A$. On the other side, $D$ is compact and $f$ continuous so the function $f$ is also bounded : $\|f\|_\infty\leq M$. Finally, because $\|.\|_\infty$ is norm or more precisely due to the triangle inequality, it implies that $|\|f_k\|_\infty-\|f\|_\infty|\leq \|f_k-f\|_\infty$. Hence you can bound $\|f_k\|_\infty$ by a constant which does not depend on $k$.2012-12-09

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