Prove that: {$f_n$} converges unifromly to $f$ on $E$ as $n \rightarrow \infty$ if and only if $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$

Question

Prove that: {$f_n$} converges unifromly to $f$ on $E$ as $n \rightarrow \infty$ if and only if $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$

I don't understand the approach for the Forward proof.

My approach is:

By definition, for $\epsilon > 0$, there exist $N \in \mathbb{N}$ such that $|f_n(x) - f(x)| < \epsilon$; hence, the smallest upper bound should also be less than the $\epsilon$ (i.e., ${sup}_{x \in E}|f_n(x)-f(x)| < \epsilon$). Thus, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.

However, the solution uses $\frac{\epsilon}{2}$ as the following:

$|f_n(x)-f| \leq {sup}_{x \in E}|f_n(x)-f(x)| < \epsilon $

${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2}$

${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2} < \epsilon$

Hence, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.

Why?

Answer 1

Supremum preserves only weak inequalities, not strict inequalities. If all elements of a set $A$ are smaller than $\epsilon$ it only implies that $\sup (A)\leq\epsilon$, and not necessary $\sup (A)<\epsilon$. That's why they use $\frac{\epsilon}{2}$. If all the elements of $A$ are less than $\frac{\epsilon}{2}$ then $\sup(A)\leq\frac{\epsilon}{2}<\epsilon$. This way we indeed get the strict inequality $\sup(A)<\epsilon$.

Not that it really matters. If you prove that for each $\epsilon>0$ we eventually have $||f_n-f||\leq\epsilon$ then it proves that $||f_n-f||\to 0$. But I guess they wanted to prove it with strict inequality.

Answer 2

You have $|f_n(x) - f(x)| < \varepsilon$. But this only promises $\sup_x |f_n(x) - f(x)| \leq \varepsilon$.

Just talking about points for a second, every point in $(0,1)$ is ${}<1$, but the supremum of that set is $1$.

Answer 3

First, we define what we mean by uniform convergence.

Let $E$ be any non-empty set, and let $(X, d)$ be any metric space. For each $n \in \mathbb{N}$, let $f_n \colon E \to X$ be a function, and let $f \colon E \to X$ be any given function. Then the sequence $\left( f_n \right)_{n \in \mathbb{N} }$ of functions converges uniformly to the function $f$ on set $E$ if, for every real number $\varepsilon > 0$, there exists a natural number $N = N(\varepsilon)$ [This means that this $N$ in general depends on our choice of the $\varepsilon$ (alone).] such that $$ d \left( f_n(s), f(s) \right) < \varepsilon \tag{1} $$ for all $s \in E$ and for any natural number $n > N$.

Then the result I think can be stated in full generality as follows:

Let $E$ be any non-empty set, and let $(X, d)$ be any metric space. For each $n \in \mathbb{N}$, let $f_n \colon E \to X$ be a function, and let $f \colon E \to X$ be any given function. Then the sequence $\left( f_n \right)_{n \in \mathbb{N} }$ of functions converges uniformly to the function $f$ on $E$ if and only if $$ \lim_{n \to \infty} \left\lVert f_n - f \right\rVert_u = 0. $$ For each $n \in \mathbb{N}$ we define $\left\lVert f_n - f \right\rVert_u$ as follows: $$ \left\lVert f_n - f \right\rVert_u \colon= \sup \left\{ \, d \left( f_n(s), f(s) \right) \, \colon \, s \in E \, \right\}, \tag{2} $$ and the last supremum is to be considered in the extended real number system.

Proof:

First, suppose that the sequence $\left( f_n \right)_{n \in \mathbb{N}}$ converges uniformly to the function $f$ on set $E$.

Let us arbitrarily choose a real number $\varepsilon > 0$. Then for the real number $\frac{\varepsilon}{2} > 0$, we can choose a natural number $N$ such that $$ d \left( f_n(s), f(s) \right) < \frac{\varepsilon}{2} $$ for all $s \in E$ and for all natural numbers $n > N$. Refer to (1) above. Here we have simply applied our definition above of uniform convergence with $\frac{\varepsilon}{2}$ in place of $\varepsilon$.

Thus for any natural number $n > N$, the set $$ \left\{ \, d \left( f_n(s), f(s) \right) \, \colon \, s \in E \, \right\} \tag{3} $$ is a non-empty bounded above subset of the set $\mathbb{R}$ of real numbers, the real number $\frac{\varepsilon}{2}$ being one of the upper bounds of this set. Therefore we can conclude that, for any natural number $n > N$, we have $$ \sup \left\{ \, d \left( f_n(s), f(s) \right) \, \colon \, s \in E \, \right\} \leq \frac{\varepsilon}{2}, $$ that is, $$ \left\lVert f_n - f \right\rVert_u \leq \frac{\varepsilon}{2}; $$ [Refer to (2) above.] but since the set in (3) above consists only of non-negative real numbers, therefore we can conclude that, for any natural number $n > N$, the following holds: $$ 0 \leq \left\lVert f_n - f \right\rVert_u \leq \frac{\varepsilon}{2}. \tag{4} $$

However, as $\varepsilon > 0$, so $$ -\varepsilon < 0 \ \mbox{ and } \ \frac{ \varepsilon}{2} < \varepsilon, $$ and therefore (4) implies that $$ -\varepsilon < \left\lVert f_n - f \right\rVert_u \leq \varepsilon $$ for any natural number $n > N$. Thus for any natural number $n > N$, we have $$ \big\lvert \, \left\lVert f_n - f \right\rVert_u \, - \, 0 \, \big\rvert < \varepsilon. \tag{5} $$ Here we have used a familiar property of real numbers and their absolute values.

But $\varepsilon$ is an arbitrarily chosen positive real number. Thus for any real number $\varepsilon > 0$, there exists a natural number $N$ such that (5) holds for any natural number $n > N$. Hence $$ \lim_{n \to \infty} \left\lVert f_n - f \right\rVert_u = 0. $$

Is this part of the proof clear enough for you?

Conversely, suppose that $$ \lim_{n \to \infty} \left\lVert f_n - f \right\rVert_u = 0. $$ Let $\varepsilon > 0$ be arbitrary. Then there exists an $N$ such that (5) above holds for any $n > N$. That is, $$ \left\lVert f_n - f \right\rVert_u < \varepsilon \tag{6} $$ for any $n > N$.

Now let $s \in E$ be arbitrary. Then for any natural number $n$, using (2) above, we can conclude that $$ d \left( f_n(s), f(s) \right) \leq \left\lVert f_n - f \right\rVert_u. $$ Finally from the last inequality and (6) above we can conclude that $$ d \left( f_n(s), f(s) \right) < \varepsilon $$ for all $s \in E$ and for any $n > N$.

Since $\varepsilon > 0$ was arbitrary, it follows that our sequence $\left( f_n \right)_{n \in \mathbb{N} }$ converges uniformly to the function $f$ on set $E$.

Hope you can get this now.