Showing that $X=\{ f \in C[a,b] : f(a) = 0 \}$ is Banach

Question

Exercise :

Show that the space $$X=\{f \in C[a,b] : f(a) = 0 \}$$ equipped with the norm $\| \cdot \|_\infty$ is a Banach space.

Attempt :

Let $\{f_n\}$ be a Cauchy sequence over $X$. Let $\epsilon >0$, then $ \exists n_0 \in \mathbb N$ such that :

$$\| f_n - f_m \|_\infty < \epsilon /2 \; \forall \; n,m \geq n_0$$ $$\Leftrightarrow \max|f_n(x) - f_m(x)| < \epsilon /2 \; \forall n, m \geq n_0 $$ $$\Rightarrow |f_n(x)-f_m(x)|<\epsilon/2 \; \forall n,m \geq n_0$$ which means that the sequence $f_n(x)$ is a Cauchy sequence of real numbers $\forall x \in [a,b]$, thus $(f_n(x))$ fonverges as $n \to \infty \; \forall x \in [a,b]$, thus $\lim f_n(x) = f(x) \; \forall x \in [a,b]$.

For $m \to \infty$ we yield :

$$|f_n(x) - f(x) | \leq \epsilon /2 < \epsilon \; \forall n \geq n_0, \; \forall x \in [a,b]$$ $$\Rightarrow \|f_n-f\| = \max|f_n(x)-f(x)|< \epsilon \; \forall n\geq n_0$$ $$\Rightarrow \lim f_n = f \; \text{over the norm} \; \| \cdot\|_\infty$$

Now, we know that if $f_n \to f$ uniformly with $(f_n)$ being a sequence of real functions, this means that $f$ is continuous, thus $f \in C[a,b]$.

Question : How does $f(a) =0$ come into the play ? How would I proceed to showing that not only $f \in C[a,b]$ but also $f \in X$ ? I know I need to show $f(a) =0$ for all these $f$s yielded by the sequences, but how ?

Answer 1

If ${\{f_n\}}_{n = 1}^{\infty}$ is a Cauchy sequence in $X$, you have shown that there exists $f \in C([a , b])$ such that $\lim_{n \to \infty} f_n(x) = f(x)$ for $x \in [a , b]$. Fix then $x = a$. Then $\lim_{n \to \infty} f_n(a) = f(a)$. But since $f_n \in X$ for all $n = 1 , 2 , \ldots$, then $f_n(a) = 0$ for all $n = 1 , 2 , \ldots$, so $$ f(a) = \lim_{n \to \infty} f_n(a) = \lim_{n \to \infty} 0 = 0\mbox{.} $$