Answer: Uniform Limit of Derivatives via Integration

Key Idea / Intuition

Uniform convergence alone is not enough to swap limits and derivatives — differentiability is a local property that requires controlling difference quotients, and uniform convergence only controls function values. However, integration is much more forgiving: uniform convergence does pass through integrals, and since a function is recovered from its derivative by the Fundamental Theorem of Calculus, we can bootstrap this to conclude differentiability when the derivatives also converge uniformly.

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Formal Proof / Solution

Part (a): Counterexample

Uniform convergence does not preserve differentiability.

Counterexample: Let $f_n(x) = \sqrt{x^2 + \tfrac{1}{n}}.$ Each $f_n$ is smooth on $[0,1]$, and $|f_n(x) - |x|| = \sqrt{x^2 + \tfrac{1}{n}} - |x| = \frac{1/n}{\sqrt{x^2+1/n}+|x|} \leq \frac{1/n}{\sqrt{1/n}} = \frac{1}{\sqrt{n}} \to 0$ uniformly. So $f_n \to f = |\cdot|$ uniformly on $[0,1]$, but $f(x) = x$ is not differentiable at $x = 0$.

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Part (b): Uniform convergence of derivatives implies differentiability of the limit

Setup. We know:

• $f_n \rightrightarrows f$ uniformly on $[0,1]$,
• $f_n' \rightrightarrows g$ uniformly on $[0,1]$.

Step 1: Integrate $f_n'$.

By the Fundamental Theorem of Calculus, for each $n$ and each $x \in [0,1]$: $f_n(x) - f_n(0) = \int_0^x f_n'(t)\, dt.$

Step 2: Pass to the limit using uniform convergence.

Since $f_n' \rightrightarrows g$ uniformly, we can pass the limit inside the integral: $\int_0^x f_n'(t)\, dt \to \int_0^x g(t)\, dt \quad \text{uniformly in } x.$

Since $f_n(x) \to f(x)$ and $f_n(0) \to f(0)$, taking $n \to \infty$ gives: $f(x) - f(0) = \int_0^x g(t)\, dt.$

Step 3: Conclude differentiability.

The right-hand side is the integral of a continuous function $g$ (uniform limit of continuous functions is continuous). By the Fundamental Theorem of Calculus: $f'(x) = g(x) \quad \text{for all } x \in [0,1].$

Thus $f$ is (continuously) differentiable and $f' = g$. $\blacksquare$

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Summary of the Trick

| Question | Answer | |---|---| | $f_n \rightrightarrows f$ implies $f$ differentiable? | No — counterexample: $\sqrt{x^2+1/n} \to |x|$ | | $f_n \rightrightarrows f$ and $f_n' \rightrightarrows g$ implies $f' = g$? | Yes — integrate $f_n'$, swap limit and integral, differentiate |

The moral: to prove a derivative statement, integrate first, then differentiate. This is a recurring theme in analysis — integration is the robust operation that tolerates limit-swapping, while differentiation is fragile.