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I am hoping for a check on my work on the question below. Comments welcomed. I'm worried I've missed something because my answer is so straight forward.

Let $(f_n)^{\infty}_{n=1}$ be a sequence of functions on an interval $I$ and let $f$ be a function on $I$. For each $n$, set

$$M_{n}=\sup \{ |f_{n}(x)-f(x)|:x \in I \}.$$

Show that $\lim_{n \to \infty} f_{n}(x) = f(x)$ uniformly on $I$ if and only if $\lim_{n \to \infty} M_n = 0$.

So, by definition, $f_n (x)$ converges uniformly to $f (x)$ if and only if

$$\lim_{n \to\infty} \left \| f_n(x)-f(x) \right \|_{\infty} = 0.$$

From what I understand, $\left \| \cdot \right \|_{\infty} $ is called the sup-norm (but I've also seen it defined as the max-norm). It is defined as

$$\left \| f_n(x)-f(x) \right \|_{\infty}= \sup \{ |f_n(x_1) - f(x_1)|,...,|f_n(x_k) - f(x_k)| \}$$

for $x_1,x_2,...,x_k \in I$. It seems this question can be answered directly from the definitions.

Assume $\lim_{n \to \infty} M_n = 0$, then we have

$$\lim_{n \to \infty} M_n = \lim_{n \to \infty} \sup \{ |f_{n}(x)-f(x)|:x \in I \}=\lim_{n \to \infty} \left \| f_n(x)-f(x) \right \|_{\infty}=0$$

and as such we have uniform convergence with $\lim_{n \to \infty} f_{n}(x) = f(x)$ on $I$.

Now, assume $\lim_{n \to \infty} M_n \neq 0$, then

$$\lim_{n \to \infty} M_n = \lim_{n \to \infty} \sup \{ |f_{n}(x)-f(x)|:x \in I \}=\lim_{n \to \infty} \left \| f_n(x)-f(x) \right \|_{\infty} \neq 0 $$

and $f_n(x)$ does not converge uniformly to $f(x)$ on $I$.

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2 Answers 2

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Generally the $\sup$-norm is defined as: $$\lVert f_n(x)-f(x)\rVert_\infty=\sup\bigl\{|f_n(x)-f(x)|:x\in I\bigr\}.$$ If you assume the interval $I$ is closed and bounded then the definition with the max is the equivalent. If you assume: $\lim_{n \to \infty} M_n = 0$ then $\forall \epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall n> N |M_n|<\epsilon$ that means $\sup\bigl\{|f_n(x)-f(x)|:x\in I\bigr\}<\epsilon$ and: $\lim_{n \to \infty} \left \| f_n(x)-f(x) \right \|_{\infty}=0$.

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Welcome to math.stackexchange. You have to enclose LaTeX with dollar signs for it to work. Please see the FAQ or check how I edited your post. –  Gibarian Mar 13 '13 at 15:29

If that's how $\lVert\cdot\rVert_\infty$ is defined for you, then you have to justify that $$\sup\bigl\{|f_n(x)-f(x)|:x\in I\bigr\}=\lVert f_n(x)-f(x)\rVert_\infty,$$ or at least, that $$\lim_{n\to\infty}\sup\bigl\{|f_n(x)-f(x)|:x\in I\bigr\}=\lim_{n\to\infty}\lVert f_n(x)-f(x)\rVert_\infty.$$ I don't think that's right, though. What are $x_1,...,x_k$? Are they arbitrary? Fixed?

Another relevant question: Are you assuming that $I$ is a closed and bounded interval?

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