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

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 ?

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 $$f(a) = \lim_{n \to \infty} f_n(a) = \lim_{n \to \infty} 0 = 0\mbox{.}$$