1.10

1. Test the following sequences of functions for pointwise/uniform convergence.

State clearly whether pointwise and uniform limits exist:

(a) \(f_n = \frac{x}{n} \ln\left(\frac{x}{n}\right), \quad x \in [2, +\infty)\)​

(b) \(f_n = \frac{1}{n} \sqrt{n^2 x^2 + 1}, \quad x \in \mathbb{R}\)

First, we prove the pointwise limit.

\(\lim_{n\to\infty}f_{n}=\lim_{n\to\infty}\frac{1}{n}\sqrt{n^{2}x^{2}+1}\Rightarrow \lim_{n\to\infty}f_{n}=\lim_{n\to\infty}\sqrt{x^{2}+\frac{1}{n^2}}=\left|x\right|\)

Then we use definition to prove the uniform convergence

For \(\epsilon > 0,\) I need to find \(N(\epsilon)\) s.t \(|f_{n}(x)-\left|x\right||<\epsilon,\forall n\geq N(\epsilon),\forall x\in\mathbb{R}\)

Thus \(\left|f_{n}-\right|x\left|\right|=\left|\sqrt{x^{2}+\frac{1}{n^2}}-\sqrt{x^{2}}\right |=\left(\sqrt{x^{2}+\frac{1}{n^2}}-\sqrt{x^{2}}\right)\cdot\frac{\sqrt{x^{2}+\frac{1}{n^2}}+\sqrt{x^{2}}}{\sqrt{x^{2}+\frac{1}{n^2}}+\sqrt{x^{2}}} =\frac{x^{2}+\frac{1}{n^2}-x^{2}}{\sqrt{x^{2}+\frac{1}{n^2}}+\sqrt{x^{2}}}=\frac{1}{\sqrt{n^{4}x^{2}+n^{2}}+\sqrt{n^{4}x^{2}}}\)

\(\frac{1}{n\sqrt{n^2x^2+1}+n^{2}\sqrt{x^2}}\)​

(c) \(f_n(x) = \frac{nx}{1 + nx^2}\)

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First, we prove the pointwise limit.

\(\lim_{n\to\infty}f_{n}=\lim_{n\to\infty}\frac{nx}{1+nx^{2}}\Rightarrow\lim_{n\to\infty} f_{n}=\lim_{n\to\infty}\frac{1}{\frac{1}{nx}+x}=\frac{1}{x}\)

Then we use definition to prove the uniform convergence

For \(\epsilon > 0,\) I need to find \(N(\epsilon)\) s.t \(|f_{n}(x)-\frac{1}{x}|<\epsilon,\forall n\geq N(\epsilon),\forall x\in\mathbb{R}\)

Thus \(|f_{n}-\frac{1}{x}|=|\frac{nx}{1+nx^{2}}-\frac{1}{x}|=|\frac{nx^{2}-1-nx^{2}}{x+nx^{3}} |=\left|\frac{1}{x+nx^{3}}\right|\)

Then we can take \(N>\frac{1}{2\varepsilon}\), it is uniform converges

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2. Determine if the following sentences are true or false:

(a)

If \(f_n \to^u f\) on \(A\) and \(f_n \to^u f\) on \(B\), then \(f_n \to^u f\) on \(A \cup B\).

(b)

If \(f_n \to^u f\) on an interval \(I\) and \(f_n\) is increasing on \(I\) for all \(n\), then \(f\) is increasing on \(I\).

(c)

If \(f_n \to f\) and \(f_n\) is bounded for all \(n\), then \(f\) is bounded.