Sequence of convex functions converges uniformly

I am working on the following problem.

Let $f_{n}: [a, b] \rightarrow \mathbb{R}$ be a sequence of convex functions. Furthermore, for each fixed $x \in [a, b]$, suppose $f(x) = \lim_{n \rightarrow \infty}f_{n}(x)$ exists and $f(x)$ is continuous on $[a, b]$. Show that $f_{n} \rightarrow f$ uniformly.

The solution seems to have been mentioned here and here. However, the solutions don't seem very satisfactory to me, so I decided to write out explicitly my own solution (partially because they start with a proof by contradiction). Can anyone check my solution to make sure it is correct? I've tried to formulate a direct proof.

Fix $\varepsilon > 0$. Since $f$ is continuous on $[a, b]$, it is uniformly continuous there, and hence there exists an $\delta > 0$ such that when $|x - y| < \delta$, $|f(x) - f(y)| <\varepsilon/10$.

Partition $[a, b]$ into $\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{m}\}$ where $\alpha_{0} = a$ and $\alpha_{m} = b$ such that $|\alpha_{i + 1} - \alpha_{i}| = \delta/2$, $i = 0, 1, 2, \ldots, m - 2$ and possibly $|\alpha_{m} - \alpha_{m - 1}| < \delta/2$.

Since for each fixed $x$, $f_{n}(x)$ converges to $f(x)$, there exists an integer $N$ such that $|f_{n}(\alpha_{i}) - f(\alpha_{i})| < \varepsilon/10$ for all $i = 0, 1, \ldots, m$ and $n \geq N$.

For $x \in [a, b]$, $x \in [\alpha_{i}, \alpha_{i + 1}]$ for some fixed $i$. Then $x = \lambda_{x} \alpha_{i} + (1 - \lambda_{x})\alpha_{i + 1}$ for $\lambda_{x}$ between 0 and 1. Therefore by convexity of $f_{n}$, for $n \geq N$ (which is independent of $x$), we have \begin{align*} f_{n}(x) - f(x) &\leq \lambda_{x} (f_{n}(\alpha_{i}) - f(x)) + (1 - \lambda_{x})(f_{n}(\alpha_{i + 1}) - f(x))\\ &\leq |f_{n}(\alpha_{i}) - f(x)| + |f_{n}(\alpha_{i + 1}) - f(x)|\\ &\leq |f_{n}(\alpha_{i}) - f(\alpha_{i})| + |f(\alpha_{i}) - f(x)| + |f_{n}(\alpha_{i + 1}) - f(\alpha_{i + 1})| + |f(\alpha_{i + 1}) - f(x)|\\ & < 2\varepsilon/5 \end{align*} where the first and third terms in the sum is $< \varepsilon/10$ since $n \geq N$ and the second and fourth terms in the sum is $< \varepsilon/10$ since $x$ is between $\alpha_{i}$ and $\alpha_{i + 1}$ and $|\alpha_{i + 1} - \alpha_{i}| \leq \delta/2$ (and hence apply uniform continuity of $f$).

We now show the other direction and consider $f(x) - f_{n}(x)$. Since $|f_{n}(\alpha_{i}) - f(\alpha_{i})| < \varepsilon/10$ for $i = 0, 1, \ldots, m$ and $n \geq N$, we may assume that $x \neq \alpha_{i}$. Fix $x \in [a, b]$, then $x \in (\alpha_{i}, \alpha_{i + 1})$ for some fixed $i$. Then there exists a $\mu_{x}$ such that $f_{n}(\alpha_{i}) \leq \mu_{x}f_{n}(x) + (1 - \mu_{x})f_{n}(\alpha_{i - 1})$. That is $$\frac{f_{n}(\alpha_{i}) - (1 - \mu_{x})f_{n}(\alpha_{i - 1})}{\mu_{x}} \leq f_{n}(x).$$ Therefore \begin{align*} f(x) - f_{n}(x) &\leq f(x) - \frac{1}{\mu_{x}}f_{n}(\alpha_{i}) + \frac{1 - \mu_{x}}{\mu_{x}}f_{n}(\alpha_{i - 1})\\ &= \frac{1}{\mu_{x}}(f(x) - f_{n}(\alpha_{i})) + \left(1 - \frac{1}{\mu_{x}}\right)(f(x) - f_{n}(\alpha_{i - 1}))\\ &\leq \frac{1}{\mu_{x}}|f(x) - f_{n}(\alpha_{i})| + \left|1 - \frac{1}{\mu_{x}}\right||f(x) - f_{n}(\alpha_{i - 1})|\\ &\leq \frac{1}{\mu_{x}}\frac{\varepsilon}{5} + \left|1 - \frac{1}{\mu_{x}}\right|\frac{\varepsilon}{5} \end{align*} where the last inequality is for $n \geq N$ and is by the same reasoning as in the $f_{n}(x) - f(x) < 2\varepsilon/5$ case.

Since $x \in (\alpha_{i}, \alpha_{i + 1})$, \begin{align*} \frac{1}{\mu_{x}} = \frac{x - \alpha_{i - 1}}{\alpha_{i} - \alpha_{i - 1}} > 1 \end{align*} and \begin{align*} \left|\frac{1}{\mu_{x}}\right| \leq \left|\frac{\alpha_{i + 1} - \alpha_{i - 1}}{\alpha_{i} - \alpha_{i - 1}}\right| \leq \frac{|\alpha_{i + 1} - \alpha_{i}| + |\alpha_{i} - \alpha_{i - 1}|}{\alpha_{i} - \alpha_{i - 1}} = 1 + \left|\frac{\alpha_{i + 1} - \alpha_{i}}{\alpha_{i} - \alpha_{i - 1}}\right| < 2. \end{align*} Therefore, we have \begin{align*} \frac{1}{\mu_{x}}\frac{\varepsilon}{5} + \left|1 - \frac{1}{\mu_{x}}\right|\frac{\varepsilon}{5} = \left(\frac{2}{\mu_{x}} - 1\right)\frac{\varepsilon}{5} \leq \frac{3\varepsilon}{5}. \end{align*} That is, we have found an $N$ (independent of $x$) such that for $n \geq N$, $f(x) - f_{n}(x) \leq 3\varepsilon/5$.

Therefore $f_{n} \rightarrow f$ uniformly.

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Convexity says $f_n(\bar{x}) \le \lambda f_n(a_i) + (1-\lambda)f_n(a_{i+1})$ -- but you cannot just pass to the absolute values in this inequality. Therefore your proof is not complete. – Jochen Aug 22 '13 at 9:09
@Jochen, Hi! Thanks for your comment, I think I have fixed the proof. – JKL629 Aug 22 '13 at 14:49
“Then there exists a $\mu_{x}$ such that $f_{n}(\alpha_{i}) \leq \mu_{x}f_{n}(x) + (1 - \mu_{x})f_{n}(\alpha_{i - 1})$”. If $i=0$ then there is no $\alpha_{i - 1}$. – Alex Ravsky Jan 19 '15 at 5:05
“where the last inequality is for $n \geq N$ and is by the same reasoning as in the $f_{n}(x) - f(x) < 2\varepsilon/5$ case”. In this case the inequality is proved only from one side. – Alex Ravsky Jan 19 '15 at 5:11