Over the given interval
I
determine the sets of uniform convergence for:
f
n
(x)=
sin(nx)
n

√

,I=R.

Question 1
Select one:

a.
f
⟶
uniform
0
on
[0,1].

b.
f
⟶
uniform
0
on
R.

c.
f
⟶
uniform
0
on
[0,∞).

d.
f
⟶
uniform
0
on

Let's analyze the sequence of functions $ f_n(x) = \frac{\sin(nx)}{\sqrt{n}} $ over the interval $ I = \mathbb{R} $ and determine the set on which it converges uniformly.

### Step 1: Pointwise Convergence
First, let's determine the pointwise limit of $ f_n(x) $ as $ n \to \infty $.

Given:
\[ f_n(x) = \frac{\sin(nx)}{\sqrt{n}} \]

As $ n $ increases, $ \sin(nx) $ oscillates between $-1$ and $1$, but the amplitude of the function is scaled by $ \frac{1}{\sqrt{n}} $. This means that:
\[ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{\sin(nx)}{\sqrt{n}} = 0 \]
for all $ x \in \mathbb{R} $.

So, $ f_n(x) $ converges pointwise to $ 0 $ on $ \mathbb{R} $.

### Step 2: Uniform Convergence
Next, we need to check whether this convergence is uniform over different intervals.

To determine uniform convergence, consider the supremum of $ |f_n(x)| $ over any interval $ A $:

\[ \sup_{x \in A} |f_n(x)| = \sup_{x \in A} \left|\frac{\sin(nx)}{\sqrt{n}}\right| = \frac{1}{\sqrt{n}} \sup_{x \in A} |\sin(nx)| \]

Since $ |\sin(nx)| \leq 1 $ for all $ x $ and $ n $, we have:
\[ \sup_{x \in A} |f_n(x)| \leq \frac{1}{\sqrt{n}} \]

As $ n \to \infty $, $ \frac{1}{\sqrt{n}} $ approaches 0. This indicates that $ f_n(x) $ converges uniformly to 0 over any bounded interval in $ \mathbb{R} $.

However, for unbounded intervals like $ \mathbb{R} $ or $ [0, \infty) $, the supremum over $ x $ could potentially involve values of $ x $ where $ \sin(nx) $ does not behave uniformly well, so we need to be more careful.

### Step 3: Analyze Options
- **Option a: $ f_n(x) $ converges uniformly to 0 on $ [0, 1] $.**
  - On the interval $ [0, 1] $, which is bounded, we can apply our earlier result. $ \sup_{x \in [0, 1]} |f_n(x)| \leq \frac{1}{\sqrt{n}} $ uniformly converges to 0 as $ n $ increases.

- **Option b: $ f_n(x) $ converges uniformly to 0 on $ \mathbb{R} $.**
  - On the entire real line, $ \sup_{x \in \mathbb{R}} |f_n(x)| $ could be problematic because $ x $ could be large, causing oscillations that do not die out uniformly. Hence, this option is less likely to be true.

- **Option c: $ f_n(x) $ converges uniformly to 0 on $ [0, \infty) $.**
  - Similar to the argument above for the entire real line, the interval $ [0, \infty) $ is unbounded, and uniform convergence over an unbounded interval is generally more challenging to establish.

- **Option d: $ f_n(x) $ converges uniformly to 0 on some unspecified set.**
  - This is too vague to be a valid answer.

### Conclusion
The correct choice is:

**a. $ f_n(x) $ converges uniformly to 0 on $ [0, 1] $.**

Would you like further details or have any questions?

---

Here are five relative questions to expand your understanding:

1. How do we determine the pointwise limit of a sequence of functions?
2. What is the difference between pointwise convergence and uniform convergence?
3. Why is uniform convergence more difficult to achieve on unbounded intervals?
4. Can you think of an example where a function converges pointwise but not uniformly?
5. How does the behavior of $ \sin(nx) $ impact the convergence properties of $ f_n(x) $?

**Tip:** Uniform convergence implies that the functions in the sequence get uniformly close to the limit function, regardless of the specific point in the domain.

Explore the uniform convergence analysis of the sequence of functions $f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ over the real line. Understand the pointwise limit and the conditions under which uniform convergence holds, with detailed mathematical reasoning.

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