Suppose $f : [0,1] \to \mathbb R$ be such that $$\displaystyle\sum_{i=1}^n |f(t_i) - f(t_{i-1})|^2 < 100$$ for any $n \in \mathbb N$ and $0 \leq t_0 < \cdots < t_n \leq 1$. Prove that $f$ is Riemann integrable on $[0,1]$.
Let $\epsilon > 0$. If we can find a partition $P$ of $[0,1]$ such that $U(P,f) - L(P,f) < \epsilon$, then we would be done. by using the fact that $b-a \leq (b-a)^2 + \frac{1}{4}$ for any real numbers $a,b$, I have proven the inequality when $\epsilon \geq 1/4$. However, I am stuck on how to prove the case for $\epsilon < 1/4$. I would greatly appreciate any help. What is the intuition on the given condition, as well as how to think of this problem?
Let $P$ be a partition of $[0,1]$. By AM-QM, $$\frac{1}{n}\sum_{i=1}^{n}|f(t_{i}) - f(t_{i-1})| \leq \sqrt{\frac{1}{n}\sum_{i=1}^{n}|f(t_{i}) - f(t_{i-1})|^{2}} <\frac{10}{\sqrt{n}}$$ Hence, $$\sum_{i=1}^{n}|f(t_{i}) - f(t_{i-1})| < 10\sqrt{n}$$ Let $P$ be the partition that partitions $[0,1]$ into intervals of length $1/n$. Choose $c_{i}$ and $c_{i}'$ such that $\sup_{x \in [t_{i-1},t_{i}]}f(x) - \frac{1}{n} < f(c_{i})$ and $\inf_{x\in [t_{i-1},t_{i}]}f(x) + \frac{1}{n} > f(c_{i}')$. We note the set of all $c_{i}$ and $c_{i}'$ form a partition of $[0,1]$. Then, \begin{align} U(f,P) - L(f,P) &= \sum_{i=1}^{n}\sup_{x\in [t_{i-1},t_{i}]} f(x)(t_{i}-t_{i-1}) - \sum_{i=1}^{n}\inf_{x\in [t_{i-1},t_{i}]} f(x)(t_{i}-t_{i-1}) \\ &=\frac{1}{n}\sum_{i=1}^{n}\left(\sup_{x\in [t_{i-1},t_{i}]}f(x)-\inf_{x\in [t_{i-1},t_{i}]}f(x)\right) \\ &\leq \frac{1}{n}\sum_{i=1}^{n}\left(|f(c_{i}) - f(c_{i}')| + \frac{2}{n}\right) \\ &<\frac{1}{n}\left(20\sqrt{n} + 2\right) \\ &\leq \frac{22}{\sqrt{n}} \end{align} We can choose $n$ appropriately given an $\varepsilon$ and it then follows that $f$ is Riemann integrable.