I'm not sure I quite follow. How do you know that $|f_n(x)-f_n(y)|$ tends to 0 as $n\to \infty$?

Also... if you want to derive a contradiction then what you need to assume is that:

There's an interval $T$ and an $\epsilon$ such that for any subinterval $T\text{'}$ of $T$, _there exists_ $x$, $y$ in $T\text{'}$ s.t. $|f(x)-f(y)|>\epsilon$. (*)

You can't assume that for all $x$, $y$ in $T\text{'}$, $|f(x)-f(y)|>\epsilon$. I'm sure you meant to write (*).

By the way, even if $x-y\to 0$ as $n\to \infty$ that doesn't necessarily guarantee that $f_n(x)-f_n(y)\to 0$. This is because we're only told that $f_n$ are each continuous - we aren't told they are equicontinuous.

OK so as for how to tackle the problem... here are a few hints.

Like you say, we want to assume that the result is false and derive a contradiction. So we have an interval $T$ and an $\epsilon >0$ such that for any subinterval $T\text{'}$ of $T$ we can find $x$, $y$ in $T\text{'}$ such that $|f(x)-f(y)|>\epsilon$. Needless to say that all the intervals mentioned here are supposed to be of positive length.

Intuitively what we know is that for any subinterval $T\text{'}$ of $T$ you can find points $x$, $y$ for which $|f(x)-f(y)|>\epsilon$. But we know that $f_n\to f$. From this you should be able to show that for sufficiently high $n$ then $|f_n(x)-f_n(y)|>\epsilon$.

(If you find it easier to prove, you can replace the $\epsilon$ here with $\epsilon /2$ or whatever - it really doesn't matter.)

Take $n$ such that this is true. Since $f_n$ is continuous at $x$ and $y$, we can find intervals $I_x$ and $I_y$ such that $|f_n(x\text{'})-f_n(x)|<\epsilon /4$ and $|f_n(y\text{'})-f_n(y)|<\epsilon /4$ for $x\text{'}$ in $I_x$ and $y\text{'}$ in $I_y$. Note that if $x\text{'}$ is in $I_x$ and $y\text{'}$ is in $I_y$ then $|f_n(x\text{'})-f_n(y\text{'})|>\epsilon /2$. This basically is because $f_n(x\text{'})$ is close to $f_n(x)$ and $f_n(y\text{'})$ is close to $f_n(y)$ but $f_n(x)$ and $f_n(y)$ are quite far apart so $f_n(x\text{'})$ can't be too close to $f_n(y\text{'})$. To show the inequality formally, use the triangle inequality.

So what do we have... In words, for any subinterval $T\text{'}$ then for sufficiently high $n$ we can find subintervals $I_x$ and $I_y$ such that $f_n$ doesn't vary much on $I_x$ or $I_y$, and the values $f_n$ takes on $I_x$ are quite far apart from the values $f_n$ takes on $I_y$.

OK so that's one idea. But how does this help? Well our strategy is going to be to try and construct a point at which $f_n$ doesn't converge - because this will then give us our contradiction.

How do we get this point? Well we can't just choose some point $x$ and hope to prove that $f_n$ doesn't converge at $x$ - that would be far too much to hope for. What we want to try and do is to ''home in'' on some point, and in doing so prove that $f_n$ can't converge here.

Well here's one idea. Pick some interval $I_1$ on which $f_1$ doesn't vary much. Now find some subinterval $I_2$ and an integer $n_2$ such that $f_{n_2}$ doesn't vary much on $I_2$ and what's more the values of $f_{n_2}$ on $I_2$ are quite far apart from the values of $f_1$ on $I_1$. Now find a subinterval $I_3$ of $I_3$ and an integer $n_3>n_2$ such that $f_{n_3}$ doesn't vary much on $I_3$, and also the values of $f_{n_3}$ on $I_3$ are far apart from the values of $f_2$ on $I_2$.

And so on. Keep on going. We have nested intervals $I_1$, $I_2$, ... and a strictly increasing sequence $n_1$, $n_2$, ... such that the values $f_{n_r}$ takes are quite far apart from the values $f_{n_{r-1}}$ takes, for each $r$. Now if you've chosen the intervals $I_n$ sensibly they will have a point of intersection $x$. But it is very clear that $f_{n_r}(x)$ cannot converge so $f_n(x)$ can't converge, which is a contradiction.

So anyway all you need is to be able to construct the intervals $I_r$ (and integers $n_r$) such that everything works out. That's where the first hint comes in handy.

Please write back if anything is unclear in this sketch.

Michael

We want to prove that if $f_n:I\to R$ are continuous functions (where $I$ is a non-empty open interval in $R$), $f_n\to f$ and $\epsilon >0$ then there is a non-empty open subinterval $I\text{'}$ of $I$ such that $x$, $y$ in $I\text{'}\Rightarrow |f(x)-f(y)|\le \epsilon$.

(It really doesn't matter whether we have $\le$ or $<$ here since you could replace $\epsilon$ with $\epsilon /2$. Also from now on all intervals will be non-empty and open - I won't bother to say this every time.)

Suppose not. Then there is $\epsilon >0$ such that for every subinterval $I\text{'}$ of $I$ we can find $x$, $y$ in $I\text{'}$ with $|f(x)-f(y)|>\epsilon$.

Claim: We can find nested intervals $I=I_0\ge I_1\ge I_2\ge \dots$ (where $A\ge B$ means the interval $A$ contains the interval $B$ and they don't share endpoints) and integers $1=n_0<n_1<\dots$ such that for each $i\ge 1$:

(i) for all $x$, $y$ in $I_i$ we have $f_{n_1}(x)-f_{n_i}(y)|<\epsilon /16$
(ii) for all $x$ in $I_{i-1}$ and $y$ in $I_i$ we have $|f_{n_{i-1}}(x)-f_{n_i}(y)|>\epsilon /8$

Set $I_0=I$. Suppose $r\ge 0$ and we can find $I_0$, $I_1$, ..., $I_r$ such that (i) and (ii) hold for $1\le i\le r$. It is enough to show we can pick $I_{r+1}\le I_r$ and $n_{r+1}>n_r$ such that (i) and (ii) hold for $i=r+1$.

From the hypothesis of the question we can find $x$, $y$ in $I_r$ such that $|f(x)-f(y)|>\epsilon$. Since $f_n(x)\to f(x)$, $f_n(y)\to f(y)$ we can find an integer $n_{r+1}>n_r$ such that $|f_n(x)-f(x)|<\epsilon /4$ and $|f_n(y)-f(y)|<\epsilon /4$. Now by the triangle inequality we have:

$|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\ge |f(x)-f(y)|$

so

$\epsilon /4+|f_{n_{r+1}}(x)-f_{n_{r+1}}(y)|+\epsilon /4>\epsilon$

and

$|f_{n_{r+1}}(x)-f_{n_{r+1}}(y)|>\epsilon /2$

Pick any $z$ in $I_r$. We cannot have both $|f_{n_{r+1}}(x)-f_{n_r}(z)\le \epsilon /4$ and $|f_{n_{r+1}}(y)-f_{n_r}(z)|\le \epsilon /4$ since we'd then get by the triangle inequality:

$|f_{n_{r+1}}(x)-f_{n_{r+1}}(y)|\le |f_{n_{r+1}}(x)-f_{n_r}(z)|+|f_{n_r}(z)-f_{n_{r+1}}(y)|\le \epsilon /4+\epsilon /4=\epsilon /2$

a contradiction.

So we have at least one of $|f_{n_{r+1}}(x)-f_{n_r}(z)|>\epsilon /4$ and $|f_{n_{r+1}}(y)-f_{n_r}(z)|>\epsilon /4$.

WLOG $|f_{n_{r+1}}(x)-f_{n_r}(z)|>\epsilon /4$.

Since $f_{n_{r+1}}$ is continuous at $x$ and $x$ is contained in $I_r$, we can find an interval $I_{r+1}$ of $I_r$ such that for all $x\text{'}$ in $I_{r+1}$ we have $|f_{n_{r+1}}(x\text{'})-f_{n_{r+1}}(x)|<\epsilon /16$. WLOG $I_{r+1}$ doesn't share any endpoint with $I_r$.

For any $z\text{'}$ in $I_r$ we have $|f_{n_r}(z\text{'})-f_{n_r}(z)|<\epsilon /16$ (since we agreed (i) holds for $i=r$). By the triangle inequality

$|f_{n_{r+1}}(x)-f_{n_{r+1}}(x\text{'})|+|f_{n_{r+1}}(x\text{'})-f_{n_r}(z\text{'})|+|f_{n_r}(z\text{'})-f_{n_r}(z)|\ge |f_{n_{r+1}}(x)-f_{n_r}(z)|$

so:

$\epsilon /16+|f_{n_{r+1}}(x\text{'})-f_{n_r}(z\text{'})|+\epsilon /16>\epsilon /4$

hence:

$|f_{n_{r+1}}(x\text{'})-f_{n_r}(z\text{'})|>\epsilon /8$

for any $x\text{'}$ in $I_{r+1}$ and $z\text{'}$ in $I_r$.

So (ii) holds for $i=r+1$. (i) certainly also holds for $i=r+1$. So we have the claim.

Now pick intervals $I_i$ and integers $n_i$ as in the claim. Let $I_i=(l_i,r_i)$. Then $l_i$ is a strictly increasing sequence and is bounded above by $r_1$, so $l_i\to l$ for some real $l$.

Since, for any $i$, the sequence $i_{i+1}$, $l_{i+2}$, ... is contained in $I_{i+1}$, $l$ is contained in $[l_{i+1},r_{i+1}]$ which is contained in $(l_i,r_i)=I_i$. Hence $l$ is a point in the intersection of $I_i$.

Now condition (ii) shows that for each $i$, $|f_{n_{i-1}}(l)-f_{n_i}(l)|>\epsilon /8$.

But if $f_n(l)$ was a convergent sequence then there would exist $N$ such that $n$, $n\text{'}>N\Rightarrow |f_n(l)-f_{n\text{'}}(l)|<\epsilon /8$. If we pick $n_r>N$ then $n_{r+1}>N$ so $|f_{n_r}(l)-f_{n_{r+1}}(l)|<\epsilon /8$ which is a contradiction.