2
$\begingroup$

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces, and let $\{f_n\}$ be a sequence of functions $f_n: X \rightarrow Y$. For any function $f : X \rightarrow Y$ the following are equivalent:

(i) $\{f_n\}$ converges uniformly to $f$

(ii) $\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \rightarrow 0$ as $n$ $\rightarrow \infty$

Proof: (i) $\Rightarrow$ (ii): Assume $\{f_n\}$ converges uniformly to $f$. $\forall \epsilon > 0$ $\exists N \in \mathbb{N}$ s.t. $d_Y(f_n(x),f(x)) < \epsilon$ $\forall x \in X$ and $\forall n \geq N$. This means that $\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \leq \epsilon$ $\forall n \geq N$, and since $\epsilon$ is arbitrary, this implies $\sup\{d_Y(f_n(x),f(x)) | x \in X\} \rightarrow 0$

I don't understand why there is $\leq$ and not $<$ in $\sup\{d_Y(f_n(x),f(x)) \ | \ x \in X\} \leq \epsilon$. Does it follow from the fact that a supremum is the smallest element greater than or equal to the others?

$\endgroup$

1 Answer 1

3
$\begingroup$

If $A\subseteq \mathbb{R}$ is such that any $a\in A$ satisfies $a<b$ then $\sup A \leq b$.

The weak inequality is necessary. For instance if $A=(0,1)$, then any $a\in A$ satisfies $a<1$ however $\sup A = 1$.

This should explain the weak inequality in your question. However since $\varepsilon$ is arbitrary it doesn't really change anything.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .