Does this proof of the inequality using equivalent norms $ \vert \int^{1}_{0} f(t)dt \vert \leq \int^{1}_{0} \vert f(t) \vert dt$ work? Given a continuous function $f: [0,1] \rightarrow \mathbb{R}^{n}$, there is the standard result where we have
$$ \Bigg| \int^{1}_{0} f(t)dt \Bigg| \leq \int^{1}_{0} \vert f(t) \vert dt$$
Here $\vert \cdot \vert$ is taken to be the Euclidean norm. The proof that I have seen of this usually involves the Cauchy Schwarz inequality. I was wondering if we could also use the fact that the all norms on $\mathbb{R}^{n}$ are equivalent and then use the 1-norm 
$\| x \|_{1}= \vert x_{1} \vert + \cdots + \vert x_{n} \vert$
Using the $1$-norm, we have that 
\begin{align*}
\Bigg\| \int^{1}_{0} f(t) dt \Bigg\| &=  \Bigg\|\left(\int^{1}_{0} f_{1}(t) dt, \int^{1}_{0} f_{2}(t) dt, \ldots, \int^{1}_{0} f_{n}(t) dt\right)  \Bigg\| \\
&=\Bigg| \int^{1}_{0} f_{1}(t) dt \Bigg| + \cdots \Bigg| \int^{1}_{0} f_{n}(t) dt  \Bigg|\\
&\leq  \int^{1}_{0} |f_{1}(t)| dt  + \cdots  \int^{1}_{0} | f_{n}(t)| dt  \\
&= \int^{1}_{0} \| f(t) \|_{1} dt
\end{align*}
We have just established that 
$$ \Bigg\| \int^{1}_{0} f(t)dt \Bigg\|_{1} \leq \int^{1}_{0} \| f(t) \|_{1} dt$$
The last line follows from when we take for granted that we know the result when $n=1$.
The only thing that bothers me about this, is that saying $\| \cdot \|_{1}$ is equivalent to $\vert \cdot \vert$ is the same thing as saying that there exists constants $c, c'$ where 
$$c\vert \cdot \vert \leq \| \cdot \|_{1} \leq c' \vert \cdot \vert$$
These constants are what make the proof sit uneasy with me, I was wondering if this was enough to consider this proof as false, or is there a way around it.
 A: The argument you give for the 1-norm looks fine to me.  
However, the statement "all norms on $\mathbb{R}^n$ are equivalent" will only let you conclude that there exists a constant $C$ such that $\Bigg| \int^{1}_{0} f(t)dt \Bigg| \leq C \int^{1}_{0} \vert f(t) \vert dt$.  It won't give you $C=1$.
Also, if you thought you were avoiding the Cauchy-Schwarz inequality, you're really not: you need it to prove that the Euclidean norm is in fact a norm!
A: Another approach:  Let $\|\cdot\|$ be any norm on $\Bbb R^n$. Partition $[0,1]$ into $m$ equally spaced subintervals and form the corresponding Riemann sums. By the triangle inequality and the homogeneity of the norm,
$$
\Big\|\sum_{k=0}^{m-1}f(k/m){1\over m}\Big\|
\le\sum_{k=0}^{m-1}\|f(k/m)\|{1\over m}.
$$
As $m\to\infty$ the (vector-valued) Riemann sums converge (component-wise) to $\int_0^1 f(t)\,dt$. Likewise for the real-valued Riemanns sum on the right. Therefore the above inequality yields the desired inequality in the limit as $m\to\infty$.
A: The inequality 
$$
\left\| \int_0^1 f(t)\, dt\right\| \le \int_0^1 \|f(t)\|\, dt $$ 
actually holds for any integrable function $f\colon [0, 1]\to E$ where $E$ is a Banach space. This is one of those results that are obvious if one chooses the right way to define objects. Following Lang ("Real and Functional Analysis", "Undergraduate Analysis"), one might define such an integral on step functions first: 
$$\tag{1}
\int_0^1 \sum_{j=1}^n c_j \boldsymbol{1}_{[a_j, a_{j+1})}(x)\, dx\overset{\rm def}{=}\sum_{j=1}^n c_j (a_{j+1}-a_j),\qquad c_j\in E$$
then observe that the triangle inequality gives 
$$
\left\|\int_0^1  \sum_{j=1}^n c_j \boldsymbol{1}_{[a_j, a_{j+1})}(x)\, dx \right\| \le \int_0^1\left\| \sum_{j=1}^n c_j \boldsymbol{1}_{[a_j, a_{j+1})}(x)\right\|\, dx\le \sup\left( \left\|\sum_{j=1}^n c_j \boldsymbol{1}_{[a_j, a_{j+1})}(x) \right\|\, :\, x\in [0,1]\right).$$
The first inequality is exactly (a special case of) the sought one. The second implies that the linear operator $\int_0^1$ defined in (1) is continuous on the space of step functions equipped with uniform convergence. Therefore, one can extend it to the whole completion of that space, which happens to contain all continuous and piecewise continuous functions. This is an alternative definition of the Riemann integral. 
This construction makes it obvious that the inequality 
$$
\left\| \int_0^1 f(x)\, dx\right\| \le \int_0^1 \|f(x)\|\, dx$$
holds for all piecewise continuous functions (at least), no matter which norm one chooses on $E$. Specializing this to $\mathbb{R}^n$ one obtains an answer to the present question.
This explanation is hasty; I highly recommend the aforementioned books for a more thorough treatment. 
