Show that linear functional $L(f) = \int_0^1 f(x) dx$ is continuous Let $(C[0,1], d_1)$ be a metric space of all continuous functions $f:[0,1] \to \mathbb{R}$, $d_1$ is the $L_1$ metric
$$d_1(f,g) = \int\limits_0^1 |f(x) - g(x)| dx$$

Show that linear functional $L(f) = \int\limits_0^1 f(x) dx$ is
  continuous

I'm not sure what would be the easiest way to prove this claim, I feel like I am on the right track

Proof Attempt:


*

*Recall $L$ is continuous if for all convergent sequence $(f_n), f_n
   \to f$ as $n \to \infty \implies L(f_n) \to L(f)$ as $n \to \infty$
Let $(f_n)$ be a convergent sequence in $(C[0,1], d_1)$, we wish to
show that $L(f_n) \to L(f)$ as $n \to \infty$

*$f_n \to f$ if $\forall \epsilon >0, \exists N \in \mathbb{N}$, such
that $\forall x \in [0,1], \forall n \geq N, d_1(f_n(x), f(x)) <
   \epsilon$
By definition, $d_1(f_n(x), f(x)) < \epsilon \implies \int\limits_0^1
   |f_n(x) - f(x)| dx < \epsilon$

*If $L(f_n) \to L(f)$ as $n \to \infty$ then $|L(f_n) - L(f)| = |\int\limits_0^1 f_n(x) - f(x) dx| < \epsilon$

*But $|\int\limits_0^1 f_n(x) - f(x) dx|  \leq \int\limits_0^1 |f_n(x) - f(x)| dx $ by triangle inequality, and $\int\limits_0^1 |f_n(x) - f(x)| dx  < \epsilon$

*Therefore,  $L(f_n) \to L(f)$ as $n \to \infty$
Does this look correct?
 A: Your proof is correct. An easier way to prove the claim is to notice that $d_1$ is the metric induced by the norm:
$$\|f\|_1 = \int_0^1 |f(t)|dt$$
Note that $L:(C([0,1]), d_1) \to \Bbb R$ is continuous iff $L: (C([0,1]), \| \cdot \|_1) \to \Bbb R$ is continuous.
Hence, to prove $L$ is continuous, it is sufficient to find a constant $C$ such that $|Lf| \le C \|f\|_1$ for all $f \in C([0,1])$. But this is obvious:
$$\left| L(f) \right| \le \int_0^1 |f(t)|dt = \|f\|_1$$
A: Your proof looks correct.  However, in the second to last bullet point,

But $|\int\limits_0^1 f_n(x) - f(x) dx|  \leq \int\limits_0^1 |f_n(x) - f(x)| dx$ by the triangle inequality, and $\int\limits_0^1 |f_n(x) - f(x)| dx  < \epsilon$.

While this is somewhat of an integral generalization of the triangle inequality, most would not say it is so.  I would just omit mentioning the triangle inequality, and just say 

But $|\int\limits_0^1 f_n(x) - f(x) dx|  \leq \int\limits_0^1 |f_n(x) - f(x)| dx<\epsilon$.

A: Let's apply the following result:
Theorem
For the linear operator $T: X \rightarrow X$ the following properties are equivalient: 
(1): $T$ is bounded 
(2): $T$ is continuous
(3): $T$ is continuous at zero
So in order to show that $L(f) = \int_{0}^{1}{f(x)dx}$ is a continuous linear operator $(C[0,1], d) \rightarrow (C(0,1), d)$ it is sufficient to check that it is bounded. Indeed, $$||L|| = \sup_{f: ||f|| \
\leq 1}{|L(f)|} = \sup_{f: ||f|| \leq 1}{\int_{0}^{1}{f(x)dx}} = \sup_{f: ||f|| \leq 1}{||f||}=1$$
