$f$ uniformly continuous, $f_n(x) = f(x + 1/n)$. Show $f_n$ uniformly converges to $f$. Why is uniform continuity needed?

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be uniformly continuous on $\mathbb{R}$ and let $f_n(x) = f(x + \frac{1}{n})$ for $x \in \mathbb{R}$. Show that $f_n$ converges uniformly to $f$.

I have seen many proofs of this online, but I am wondering if the requirement that $f$ be uniformly continuous is necessary. Could we weaken this to continuity?

Suppose $f$ is continuous. Then

$\forall \epsilon > 0, \exists \delta > 0$ s.t. whenever $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$

I want to show $\forall \epsilon > 0 \ \exists N \in \mathbb{N}$ s.t. $\forall n \geq N, \forall x \in \mathbb{R}, |f_n(x) - f(x)| < \epsilon$

Now, consider $g_n(x) = x - \frac{1}{n}$ and $g(x) = x$. Choose $N = \frac{1}{\delta}$. Then,

$|g_n(x) - g(x)| = |x + \frac{1}{n} - x| = |\frac{1}{n}| < \frac{1}{N} < \delta$

Now, here is where the other proofs use uniform continuity. Since $\delta$ could depend on $x$, uniform continuity would be needed to ensure $N$ does not depend on $x$. However, let's start from a different approach.

I know $g_n(x) = x + \frac{1}{n}$ converges uniformly to $g(x) = x$. In case this needs justification, forget about the $\delta$ above for a moment.

I want to show $\forall \rho > 0, \exists N_{1} \in \mathbb{N}, \forall n \geq N_{1}, \forall x \in \mathbb{R}, |g_n(x) - g(x)| < \rho$.

Fix $\rho > 0$ Choose $N = \frac{1}{\rho}$. Then, $|g_n(x) - g(x)| = |x + \frac{1}{n} - x| = |\frac{1}{n}| < \frac{1}{N} < \rho$

Therefore, $g_n$ is uniformly convergent. This $\rho$, unlike $\delta$ from before, will definitely not depend on $x$ due to uniform convergence of $g_n$. Therefore, I can take $\delta = \rho$ in the definition of continuity, and this choice of $\delta$ will work because indeed $|g_n(x) - g(x)| < \rho = \delta$. I will also choose $N = N_{1}$. This would imply

$|f(x_n) - f(x)| < \epsilon$

and then by substitution,

$|f(x_n) - f(x)| = |f(x + \frac{1}{n}) - f(x)| < \epsilon$

Therefore, $f_n$ converges uniformly to $f$.

Considering how many times this question has been asked, I am assuming I went wrong somewhere. Does anyone see where my "proof" without uniform continuity would fall apart?

This is different from this question If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous? as this one starts with $f_n$ converging uniformly. I am trying to start from the other side.

Answer 1

Let me restate my comment as an answer: you should follow through the logic using a concrete example like $f(x) = x^2$ to see exactly where the argument fails.

Answer 2

If $f$ is only continuous and not uniformly continuous, then $\delta$ needs to depend on $x$ and $\epsilon$. If $\delta$ doesn't depend on either of them, then you cannot use it in the definition of $f$ being continuous. You've created a $\delta$ that doesn't depend on either of $x$ or $\epsilon$, so it cannot be applied to the definition.

In other words, your "proof" falls apart right here:

Therefore, I can take $\delta = \rho$ in the definition of continuity,

[...]

I will also choose $N = N_{1}$. This would imply

$|f(x_n) - f(x)| < \epsilon$

The "This would imply" isn't true, because the definition of continuity gives you a $\delta$. It cannot be chosen.

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An addendum: After a lengthy comment thread, it has become apparent to me that the issue is one of mixing up the semantics of the existential quantifier $\exists$. I'll try to be as explicit as possible using the logical syntax to guide us:

• Suppose we want to show that $f$ is continuous for an appropriately given $f$ (note this is not the context of the current problem). That is, we want to show $$\forall \varepsilon > 0, \forall x, \exists \delta > 0, \forall y, \Big(|x-y| < \delta \longrightarrow |f(x)-f(y)| < \varepsilon\Big)$$ To show this, we let $\varepsilon > 0$ be arbitrary, then we let $x$ be arbitrary, and then we find some $\delta > 0$ such that $|f(x)-f(y)| < \varepsilon$ for any $y$ such that $|x-y| < \delta.$

Much of the work in proving $f$ is continuous, therefore, comes from finding an appropriate value/formula for $\delta$. In a sense, we must choose $\delta>0$ (often as a function of $\epsilon$ and $x$) specifically so that the definition works.

• Suppose now that we are given that $f$ is continuous and we want to prove something else. Then how can we utilize the given truth of $$\forall \varepsilon > 0, \forall x, \exists \delta > 0, \forall y, \Big(|x-y| < \delta \longrightarrow |f(x)-f(y)| < \varepsilon\Big)?$$ This time, we may choose any $\varepsilon > 0$ we want and choose any $x$ we want (sometimes these may be arbitrary and other times they may be very particular values; it all depends on what you're trying to prove). However, we cannot choose $\delta > 0$. Ultimately, all the assumption that $f$ is continuous tells us is that there is some $\delta > 0$ that works for the given choice of $\varepsilon$ and $x$ without telling us anything about it's value. Therefore, we must let $\delta > 0$ be fixed but unknown to us; all we know is that if $|x-y| < \delta$, then we'd have $|f(x)-f(y)| < \varepsilon$ (and how this applies to the proof would depend on what we're trying to prove).

This is the context we find ourselves in for the problem being asked. In particular, as mentioned above, we may choose $\varepsilon > 0$ (in the usual proof, this is arbitrary but chosen as part of the definition of "$f_n$ converges uniformly to $f$") but we cannot choose $\delta > 0$.

As a sidenote, even though we cannot choose $\delta$ in this type of situation, we can still derive some specific properties for it. For instance, once we have the unknown $\delta > 0$, we can take any smaller value and it'll still work. This is how we can, for instance, replace $\delta$ with the minimum of $\delta$ and $1/2$ in order to assume $0 < \delta \leq 1/2$, but even in this case, we're almost always dealing with a fixed but unknown value for $\delta$.

Answer 3

"Suppose $f$ is continuous. Then

$\forall \epsilon > 0, \exists \delta > 0$ s.t. whenever $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$"

Here if you only assume continuous (Not UC), then your $\delta$ might depend on the point you choose, for example if you study the continuity at some point $x\in\mathbb{R}$, then $\delta=\delta_\epsilon(x)$. Hence your $N$ depends on $x$, so you can't say $f_n\rightarrow f$ uniformly. Because UC requires your $N$ is independent of the choices for point $x$.

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Proof for OP with UC condition:

$f(x)$ is uniformly continuous, which means

$$\forall \epsilon>0, \exists \delta_\epsilon, \forall x,y, ~|x-y|<\delta_\epsilon\rightarrow|f(x)-f(y)|<\epsilon \tag{*}$$

Now, take $N_\epsilon=\lfloor\frac{1}{\delta_\epsilon}\rfloor+1$,

$$~\forall n>N_\epsilon \Rightarrow n> \lfloor\frac{1}{\delta_\epsilon}\rfloor+1\ge\frac{1}{\delta_\epsilon}\Rightarrow \frac{1}{n}<\delta_\epsilon$$

$$\Rightarrow \left|\left(x+\frac{1}n\right)-x\right|=\frac{1}{n}<\delta_\epsilon$$

$$ ~|f_n(x)-f(x)|=\left|f\left(x+\frac{1}{n}\right)-f(x)\right|<\epsilon~~~~~~\text{by Eq.}(*)$$

Therefore, $f_n\rightarrow f$ uniformly.