Proving one metric on $\mathcal{C}([0,1], \mathbb{R})$ dominates another

Here are the definitions I am working with.

Define a metric on $C([0,1], \mathbb{R})$, the set of continuous functions on $[0,1]$, by $$ \rho_1 (f_1, f_2) = \max\limits_{t \in [0,1]} |f_1 (t) - f_2 (t)|. $$ We define an alternative metric on the same set by $$ \rho_2 (f_2, f_2) = \int_{t \in [0,1]} |f_1 (t) - f_2 (t)|. $$

The definition of one metric dominating another I'm working with is as follows.

Let $X$ be a set equipped with two metrics $\rho'$ and $\rho''$. We say that $\rho''$ dominates $\rho'$ if there exists a constant $C$ such that $$ \rho'(x_1, x_2) \leq C \rho''(x_1, x_2). $$

Though the definition does not say, I think it is implicit that there must exist a $C \geq 0$ that holds for all $x_1, x_2 \in X$.

The problem I'm trying to solve is then:

Show that the metric $\rho_1$ dominates $\rho_2$, but not vice versa.

For the first part, it is clear that I need to show that there exists a $C \geq 0$ such that for all $f_1, f_2 \in C([0,1], \mathbb{R})$, $\rho_1 (f_1, f_2) \leq C \rho_2 (f_2, f_2)$. How to show that the opposite is not true is not immediately clear to me, unless it boils down to negating quantifiers. I'm not sure if it's possible to prove both of these facts simultaneously.

Here is my attempt.

Let $f_1, f_2 \in C([0,1], \mathbb{R})$. By the definition of the Riemann integral, we have: \begin{align*} \rho_2 (f_1, f_2) & = \int_{t \in [0,1]} |f_1 (t) - f_2 (t)| \\ & = \lim\limits_{n \to \infty} \sum\limits_{k=1}^n |f_1 (x_k) - f_2 (x_k)| \\ & \leq \lim\limits_{n \to \infty} \sum\limits_{k=1}^n \max\{|f_1 (x_k) - f_2 (x_k)| \} \\ & = n \max\{|f_1 (x_k) - f_2 (x_k)| \} \\ & = n \rho_1 (f_1, f_2). \end{align*} I'm not sure if this is fully rigorous. What I've tried to do is to use the fact that if $|f_1 (t) - f_2 (t)|$ is integrable (I assume this is a byproduct of the functions being continuous, a different of continuous functions being continuous, and the absolute value being continuous), then any Riemann sum will converge, in the limit as $n \to \infty$, to the true value of the integral, so I've chosen an arbitrary partition.

For the reverse direction, I believe I would need to show that for all $C \geq 0$, there exists $x_1, x_2 \in X$ such that $$ \rho_2 (f_2, f_2) > \rho_1 (f_1, f_2). $$ I'm not sure how to start this direction, though. I don't believe this follows directly from the above, though what I want to say is, "the above is surely not, in general, an equality, so this statement must be true."

Your deifnition of Riemann integral is wrong. Let $(x_k)$ be a partition of $[0,1]$. Then, denoting by $\delta$ the norm of the partition we have

$$\rho_2 (f_1, f_2) = \int_{t \in [0,1]} |f_1 (t) - f_2 (t)| \\$$ $$= \lim\limits_{\delta \to 0} \sum\limits_{k=1}^n |f_1 (x_k) - f_2 (x_k)|(x_{k+1}-x_k) \\ \leq \max\{|f_1 (x) - f_2 (x)| \}$$ since $\sum (x_{k+1}-x_k)=1$.

There is no inequality in the reverse direction: Let $f_n(x)=nx$ for $0 \leq x \leq \frac 1n $ and $1$ for $x>\frac 1 n$. USe the fact that $\int_0^{1} |f_n(x)| \to 0$ and $\sup |f_n(x)|=1$ for all $n$ to see that there is no inequality in the reverse direction.