📂Analysis

Uniform Convergence of a Sequence of Functions and Differentiability

Theorem¹

$$ f^{\prime} (x) = \lim_{n \to \infty} f_{n}^{\prime} (x) \quad a \le x \le b $$

Explanation²

The result of the theorem can be summarized as “the limit of the derivatives is equal to the derivative of the limits”. In other words, it is possible to interchange the limit symbol and the differentiation symbol.

$$ \dfrac{d}{dx} \lim\limits_{n \to \infty} f_{n} (x) = \lim\limits_{n \to \infty} \dfrac{d}{dx} f_{n} (x) \quad a \le x \le b $$

The reason for considering uniform convergence of a sequence of functions concerning differentiation is, firstly, the pointwise convergence does not preserve the differentiability (Counterexample 1). Secondly, even if $f_{n} \to f$ holds and $f$ is differentiable, $f_{n}^{\prime} \to f^{\prime}$ might not hold (Counterexample 2).

Counterexample 1

Pointwise convergence of a sequence of differentiable functions $f_{n}$ to $f$ does not guarantee that $f$ is differentiable.

Proof

Function $f_{n}(x) = x^{n}$ is differentiable in $[0, 1]$. Define function $f$ as follows.

$$ f (x) = \begin{cases} 0 & \text{if } 0 \le x \lt 1 \\ 1 & \text{if } x = 1 \end{cases} $$

Then, $f_{n}(x)$ converges pointwise to $f(x)$ at every point $x \in [0, 1]$. However, it is clear that $f$ is not differentiable at $x = 1$.

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Counterexample 2

In the interval $[0, 1]$, $f_{n} \to f$, but there exist differentiable functions $f_{n}$ and $f$ such that

$$ \lim\limits_{n \to \infty} f_{n}^{\prime} (x) \ne \left( \lim\limits_{n \to \infty} f_{n}(x) \right)^{\prime} \quad \text{ for } x=1 $$

Proof

Assume $f_{n}(x) = \dfrac{x^{n}}{n}$ and $f(x) = 0$. Then, in the interval $[0, 1]$,

$$ x^{n} \to 0 \text{ and } n \to \infty \quad \text{ as } n \to \infty $$

Thus, $f_{n} \to f$ holds. However, since $f_{n}^{\prime} (x) = x^{n-1}$, it follows that $f_{n}^{\prime} (1) = 1$. Therefore,

$$ 1 = \lim\limits_{n \to \infty} f_{n}^{\prime} (1) \ne \left( \lim\limits_{n \to \infty} f_{n}(1) \right)^{\prime} = 0 $$

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Proof

Assumption: $f_{n}(x_{0}) \to f(x_{0})$ and $f^{\prime}_{n}$ uniformly converge.

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Let a small positive number $\epsilon \gt 0$ be given. By the equivalence of convergent sequences and Cauchy sequences, there exists a positive number $N$ such that the following holds due to the assumption.

$$ n, m \ge N \implies \begin{array}{l} |f_{n} (x_{0}) - f_{m} (x_{0})| \lt \dfrac{\epsilon}{2} \qquad\qquad\qquad\qquad\quad\ \ (1) \\[0.5em] \text{and} \\[0.5em] |f_{n}^{\prime} (t) - f_{m}^{\prime} (t)| \lt \dfrac{\epsilon}{2(b - a)} \quad (a \le t \le b) \qquad (2) \end{array} $$

Applying the mean value theorem to function $f_{n} - f_{m}$ along with $(2)$, we obtain the following. Given $n ,m \ge N$ and $x, t \in [a, b]$,

$$ \begin{aligned} \left| \left( f_{n}(x) - f_{m}(x) \right) - \left( f_{n}(t) - f_{m}(t) \right) \right| &= |x - t| \left| f_{n}^{\prime} (s) - f_{m}^{\prime} (s) \right| \quad (t \le s \le x) \nonumber \\ &\lt |x - t|\dfrac{\epsilon}{2(b - a)} = \dfrac{\epsilon}{2} \dfrac{|x - t|}{(b - a)} \nonumber \\ &\lt \dfrac{\epsilon}{2} \end{aligned} \tag{3} $$

From $(1)$ and $(3)$, the following inequality holds. Given $n, m \ge N$ and $x \in [a, b]$,

$$ \begin{align*} \left| f_{n}(x) - f_{m}(x) \right| &= \left| f_{n}(x) - f_{m}(x) + \big[f_{n}(x_{0}) - f_{n}(x_{0})\big] + \big[f_{m}(x_{0}) - f_{m}(x_{0}) \big] \right| \\ &\lt \left| f_{n}(x) - f_{m}(x) - (f_{n}(x_{0}) - f_{m}(x_{0})) \right| + \left| f_{n}(x_{0}) - f_{m}(x_{0}) \right|\\ &\lt \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon \end{align*} $$

The choice of $N$ is independent of $x$, so $f_{n}$ uniformly converges in the interval $[0, 1]$. Let the limits be $f(x) = \lim\limits_{n \to \infty} f_{n}(x)$ and $(a \le x \le b)$.

Now, fixing $x \in [a, b]$, define functions $\phi_{n}$ and $\phi$ as follows.

$$ \phi_{n}(t) = \dfrac{f_{n}(t) - f_{n}(x)}{t - x},\qquad \phi(t) = \dfrac{f(t) - f(x)}{t - x} \quad (x \ne t \in [a,b]) $$

Then, the following holds.

$$ \lim\limits_{t \to x} \phi_{n}(t) = \lim\limits_{t \to x} \dfrac{f_{n}(t) - f_{n}(x)}{t - x} = f_{n}^{\prime} (x) $$

Also, from the first inequality in $(3)$, we obtain the following.

$$ \left| \phi_{n}(t) - \phi_{m}(t) \right| \lt \dfrac{\epsilon}{2(b - a)} \qquad (n, m \ge N) $$

Therefore, $\phi_{n}$ uniformly converges to $t \ne x$. Thus, since $f_{n} \to f$,

$$ \phi_{n}(t) \rightrightarrows \phi(t) \quad \text{ or } \quad \lim\limits_{n \to \infty} \phi_{n}(t) = \phi (t) \quad (a \le t \le b, t \ne x) $$

Uniform convergence and continuity

If $f_{n} ⇉ f$ on a metric space $E$, then the following holds for accumulation points $x$ in $E$.

$$ \lim\limits_{t \to x}\lim\limits_{n \to \infty} f_{n}(t) = \lim\limits_{n \to \infty}\lim\limits_{t \to x} f_{n}(t) $$

Since $\phi_{n} ⇉ \phi \text{ on } [a,b]\setminus \left\{ x \right\}$, applying the above theorem to $\phi_{n}$, the following holds.

$$ \begin{align*} && \lim\limits_{t \to x}\lim\limits_{n \to \infty} \phi_{n}(t) &= \lim\limits_{n \to \infty}\lim\limits_{t \to x} \phi_{n}(t) \\ \implies && \lim\limits_{t \to x} \phi(t) &= \lim\limits_{n \to \infty} f^{\prime}_{n}(x) \\ \implies && \lim\limits_{t \to x} \dfrac{f(t) - f(x)}{t - x} &= \lim\limits_{n \to \infty} f^{\prime}_{n}(x) \\ \implies && f^{\prime}(x) &= \lim\limits_{n \to \infty} f^{\prime}_{n}(x) \end{align*} $$

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