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1.9

Let \((f_n)\) be a sequence of functions, where \(f_n : A \to \mathbb{R}\), and \(A \subseteq \mathbb{R}\).

Pointwise Convergence: \(f_n \to f\) if \(\forall \epsilon > 0, \ \exists N(\epsilon, x)\) such that \(|f_n(x) - f(x)| < \epsilon \ \forall n \geq N(\epsilon, x).\)

Uniform Convergence: \(f_n \xrightarrow{\mu} f\) if \(\forall \epsilon > 0, \ \exists N(\epsilon)\) such that \(|f_n(x) - f(x)| < \epsilon \ \forall n \geq N(\epsilon)\) and \(\forall x \in A.\)

    1. \(f_n(x) = \frac{x}{n}, \ 0 \leq x \leq 1\)​

      \(f_1(x) = x\) \(f_2(x) = \frac{x}{2}\) \(f_3(x) = \frac{x}{3}\)

      \(\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{x}{n} = x \cdot \lim_{n \to \infty} \frac{1}{n} = x \cdot 0 = 0\), then \(f_n \to 0\)

      \(f_n \xrightarrow{\mu} f = 0?\)

      For \(\epsilon > 0, I\) need to find \(N(\epsilon)\) s.t \(|f_n(x) - 0| < \epsilon, \forall n \geq N(\epsilon), \forall x \in [0, 1]\)

      \(|f_n(x) - 0| = \left|\frac{x}{n}\right| = \frac{x}{n} \leq \frac{1}{n} < \epsilon \implies \frac{1}{\epsilon} < n\)

      Choosing \(N(\epsilon) > \frac{1}{\epsilon}\), we can see that \(f_n \xrightarrow{\mu} 0\) 2. \(f_n(x) = \frac{x}{n}, \ x \in [0, \infty)\)

      \(\lim_{n \to \infty} f_n(x) = ... = 0, \ f_n \to 0\)

      We have \(f_n \xrightarrow{\mu} 0\) for \(0 < x \leq 1\)

      Let us see for \(x > 1\): \(f_n \xrightarrow{\mu} 0?\)

      \(|f_n(x) - 0| = \left|\frac{x}{n}\right| = \frac{x}{n}\)

      For \(\epsilon = 1, \ x = 2\), then \(\frac{2}{n} < 1 \implies 2 < n\), then \(N = 2\)

      \(x = 100 \implies \frac{100}{n} < 1 \implies 100 < n \implies N = 100\)

      Since \(X\) is increasing, the value of \(N\) will increase too. So we can't get an appropriate \(N(\epsilon)\) that satisfies the definition.

      Then \(f_n \cancel{\xrightarrow{\mu}} 0.\)

      Another idea: Suppose \(f_n \xrightarrow{\mu} 0\) for \(x > 1\), then you will get a contradiction. 3. \(f_{n}(x)=\frac{x^{n}}{1+x^{n}},\ x\in[0,+\infty)\)

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

      Case \(0 \leq x < 1\):

      \(\lim_{n\to\infty}\frac{x^{n}}{1+x^{n}}=\frac01=0\)

      Case \(x = 1\):

      \(\lim_{n \to \infty} \frac{1^n}{1 + 1^n} = \frac{1}{2}\)

      Case \(x > 1\):

      \(\lim_{n \to \infty} \frac{x^n}{1 + x^n} = \lim_{n \to \infty} \frac{x^n}{x^n (1 + \frac{1}{x^n})} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{x^n}} = \frac{1}{1} = 1\)

      \(f(x)=\begin{cases}0 & 0\leq x<1\\ \frac12 & x=1\\ 1 & x>1\end{cases}\)​

      \(f_n \to f\)

      \(f_n \xrightarrow{\mu} f \ ?\)

      Thm: If \(f_n\) is continuous on \(A\) and \(f_n \xrightarrow{\mu} f\) on \(A\), then \(f\) is continuous on \(A\).

      If \(f_n \xrightarrow{\mu} f \implies f\) is continuous \(\forall x \in [0, +\infty)\).

      Contradiction: \(f\) is not continuous on \(x = 1\).

      \(\therefore f_n \cancel{\xrightarrow{\mu}} f\) 4. \(f_n(x) = x^n (1 - x), \ x \in [0, 1]\)

      \(\lim_{n\to\infty}f_{n}(x)=\lim_{n\to\infty}x^{n}(1-x)=(1-x)\lim_{n\to\infty}x^{n}=\begin{cases}0 & 0\leq x<1\\ 0 & x=1\end{cases}\)

      \(f_n \to 0\) \(f_n \xrightarrow{\mu} 0 \ ?\)

      For \(\epsilon > 0, \ \text{I must find } N(\epsilon) \ \text{such that } |f_n(x) - 0| < \epsilon, \ \forall n \geq N(\epsilon), \ \forall x \in [0, 1]\)

      \(|f_n(x)| = |x^n (1 - x)|\) \(f_n(x) = x^n (1 - x)\) \(f_n'(x) = n x^{n-1} (1 - x) + x^n (-1)\)

      \(f_{n}^{\prime}(x)=nx^{n-1}-nx^{n}-x^{n}=nx^{n-1}-(n+1)x^{n}\)

      \(f_n'(x) = 0\)

      \(nx^{n-1}-(n+1)x^{n}=0\)

      \(x^{n-1} (n - (n+1) x) = 0\)

      \(n x^{n-1} (1 - \frac{(n+1)}{n} x) = 0\)

      \(\implies x = 0 \ \text{or} \ x = \frac{n}{n+1}\)

      \(f_n(0) = 0 \ \forall n\)

      \(f_n'(x) > 0\) for \(x < \frac{n}{n+1}\) \(f_n'(x) < 0\) for \(x > \frac{n}{n+1}\)

      \(\implies\) \(f_n\) has a global maximum at \(x = \frac{n}{n+1}\).

      Choose \(N(\epsilon) > \frac{1}{\epsilon}\), and the definition holds. Thus, \(f_n \xrightarrow{\mu} 0\).

      Check:

      For \(\epsilon > 0\), I must find \(N(\epsilon)\) such that \(|f_n(x) - 0| < \epsilon\) \(\forall n \geq N(\epsilon)\), \(\forall x \in [0, 1]\).

      \(|f_n(x)| = x^n (1 - x) \leq \left|f_n\left(\frac{n}{n+1}\right)\right| = \left(\frac{n}{n+1}\right)^n \left(1 - \frac{n}{n+1}\right)\).

      \(= \left(\frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}\).

      \(= \frac{1}{n+1} \cdot \left(\frac{n}{n+1}\right)^n < \frac{1}{n+1} < \frac{1}{n}\).

      To satisfy \(|f_n(x) - 0| < \epsilon\), choose \(N(\epsilon) > \frac{1}{\epsilon}\). Then \(|f_n(x) - 0| < \epsilon\) holds.

      For \(\epsilon > 0\), I must find \(N(\epsilon)\) such that \(|f_n(x) - 0| < \epsilon\) \(\forall n \geq N(\epsilon)\), \(\forall x \in [0, 1]\).

      \(|f_{n}(x)|=\left|x^{n}(1-x)\right|\leq\left|f_{n}\left(\frac{n}{n+1}\right)\right|=\left(\frac{n}{n+1}\right)^{n}\left(1-\frac{n}{n+1}\right)\).

      \(= \left(\frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}\).

      \(<1\cdot\frac{1}{n+1}<\frac{1}{n}<\varepsilon\)​ 5. \(f'(x)\) increasing and decreasing

  2. \(f_n \xrightarrow{\mu} f\) on \(A \implies f_n \to f\) on \(A\)

    Proof: \(f_n \xrightarrow{\mu} f\) on \(A\), then \(\forall \epsilon > 0\), \(\exists N(\epsilon)\) such that \(|f_n(x) - f(x)| < \epsilon \ \forall n \geq N(\epsilon)\) and \(\forall x \in A\).

    Taking \(N(\epsilon, x) = N(\epsilon)\), then the definition of pointwise convergence follows.

    Warning \(f_n \to f \cancel\implies f_n \xrightarrow{\mu} f\)

  3. Suppose that \(f_n : A \to \mathbb{R}\) is bounded on \(A\) \(\forall n\) and \(f_n \xrightarrow{\mu} f\) on \(A\). Prove that \(f\) is bounded on \(A\).

    Proof

    Definition 2 holds. In particular, for \(\epsilon = 1\), we have that \(\exists N(1)\) such that \(|f_n(x) - f(x)| < 1 \ \forall x \in A, \ \forall n \geq N(1)\)​

    \(|f_n(x)| \leq M_n, \ M_n > 0, \ \forall x \in A\)

    \(|f(x)| = |f(x) - f_n(x) + f_n(x)| \leq |f(x) - f_n(x)| + |f_n(x)|\) \(\leq1+M_{n},\ \forall n\geq N(1),\ \forall x\in A\)

    Then \(f\) is bounded on \(A\).

  4. Assume that \(f_n, g_n\) are uniformly convergent sequences of functions on \(A\).

    1. Show that \(f_n + g_n\) is uniformly convergent sequence on \(A\).

      \(f_n \xrightarrow{\mu} f\) on \(A\) and \(g_n \xrightarrow{\mu} g\) on \(A\) imply: \(\forall \epsilon > 0, \, \exists N_1\left(\frac{\epsilon}{2}\right)\) such that \(\lvert f_n(x) - f(x) \rvert < \frac{\epsilon}{2}\), \(\forall n \geq N_1\left(\frac{\epsilon}{2}\right), \, \forall x \in A\). \(\forall \epsilon > 0, \, \exists N_2\left(\frac{\epsilon}{2}\right)\) such that \(\lvert g_n(x) - g(x) \rvert < \frac{\epsilon}{2}\), \(\forall n \geq N_2\left(\frac{\epsilon}{2}\right), \, \forall x \in A\). Choose \(N(\epsilon) = \max \left\{ N_1\left(\frac{\epsilon}{2}\right), N_2\left(\frac{\epsilon}{2}\right) \right\}\).

      \(\lvert (f_n + g_n)(x) - (f + g)(x) \rvert = \lvert f_n(x) - f(x) + g_n(x) - g(x) \rvert \leq \lvert f_n(x) - f(x) \rvert + \lvert g_n(x) - g(x) \rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\).

      Thus, \(f_n + g_n \xrightarrow{\mu} f + g\). 2. Give an example to show that \(f_n \cdot g_n\) may not converge uniformly.

      \(f_n(x) = \frac{1}{x} + \frac{1}{n}, \, x \in (0, +\infty)\) \(\lim_{n \to \infty} f_n(x) = \frac{1}{x}\)

      For \(\epsilon > 0\): \(\lvert f_n(x) - \frac{1}{x} \rvert = \lvert \frac{1}{x} + \frac{1}{n} - \frac{1}{x} \rvert = \frac{1}{n} < \epsilon\)​\(\implies N(\epsilon) > \frac{1}{n}\)

      Thus, \(f_n \xrightarrow{\mu} \frac{1}{x}\).

      \(f_n(x) = g_n(x) = \frac{1}{x} + \frac{1}{n}, \, x \in (0, +\infty)\) \(f_n g_n \to \frac{1}{x^2}\)

      \(\lim_{n \to \infty} f_n(x) g_n(x) = \lim_{n \to \infty} \left( \frac{1}{x} + \frac{1}{n} \right)^2 = \lim_{n \to \infty} \left( \frac{1}{x^2} + \frac{2}{x \cdot n} + \frac{1}{n^2} \right) = \frac{1}{x^2}\)

      Let us see if \(f_n g_n \xrightarrow{\mu} \frac{1}{x^2}\).

      Suppose \(f_n g_n \xrightarrow{\mu} \frac{1}{x^2} \implies\) definition 2.

      For a particular \(\epsilon = 1\): \(\lvert f_n g_n(x) - \frac{1}{x^2} \rvert = \lvert \frac{1}{x^2} + \frac{2}{x n} + \frac{1}{n^2} - \frac{1}{x^2} \rvert = \lvert \frac{2}{x n} + \frac{1}{n^2} \rvert < 1\).

      \(\frac{2}{x n} + \frac{1}{n^2} < 1, \, \forall x \in (0, +\infty)\).

      For \(x = 1\): \(\frac{2}{n} + \frac{1}{n^2} < 1 \implies \frac{3}{n} < 1 \implies 3 < n \implies N = 3\).

      For \(x = 10\): \(\frac{2}{10 n} + \frac{1}{n^2} < 1 \implies \frac{1}{5 n} + \frac{1}{n^2} < 1 \implies \frac{6}{5 n} < 1 \implies 6 < 5 n \implies N = 2\).

      For \(x=\frac{1}{100}\), \(N\) is bigger, thus it is impossible

      Thus, \(f_n g_n \not\xrightarrow{\mu} \frac{1}{x^2}\). 3. Prove that if \(\exists M > 0\) such that \(|f_n| \leq M\) and \(|g_n| \leq M\) \(\forall n, \forall x \in A\), then \(f_n \cdot g_n\) converges uniformly on \(A\).

      \(\lvert f_n(x) \rvert \leq M\) and \(\lvert g_n(x) \rvert \leq M\).

      If \(f_n \xrightarrow{\mu} f\) and \(g_n \xrightarrow{\mu} g\), then \(f_n \to f\), \(g_n \to g\), and thus \(f_n g_n \to f \cdot g\).

      For \(\epsilon > 0\), \(\lvert f_n g_n(x) - f \cdot g(x) \rvert = \lvert f_n(x) g_n(x) - f(x) g(x) \rvert\) \(= \lvert f_n(x) g_n(x) - f(x) g_n(x) + f(x) g_n(x) - f(x) g(x) \rvert\) \(\leq \lvert f_n(x) \rvert \lvert g_n(x) - g(x) \rvert + \lvert g(x) \rvert \lvert f_n(x) - f(x) \rvert\).

      \(\leq M \cdot \frac{\epsilon}{2M} + T \cdot \frac{\epsilon}{2T} = \epsilon\).

      Thus, \(f_n g_n \xrightarrow{\mu} f \cdot g\).