I am self studying topology and I am trying to do this exercise about Vector Spaces which I have a hard time solving.
Let $\mathcal{C}([0, 1], \mathbb{R})$ be the vector space of continuous real-valued functions in the interval [0,1]. For a continuous function $f : [0, 1] \to \mathbb{R}$ we define $$ ||f||_1 = \int^{1}_{0}|f(x)| dx.$$ Now consider in $\mathcal{C}([0, 1], \mathbb{R})$ the sequences $(f_n)$ and $(g_n)$ defined by:
$$ f_n(x) = \begin{cases} 1 - nx, & \text{for } 0\leq x \leq 1/n\\ 0, & \text{for } 1/n \leq x \leq 1 \end{cases} \\ g_n(x) = \begin{cases} n-n^2x, & \text{for } 0\leq x \leq 1/n\\ 0, & \text{for } 1/n \leq x \leq 1 \end{cases} $$
Now I must examine the convergence of each sequences and determine the limit function if convergent.
In this book I am working with, I can't find any theorem that could help me out on this one so I really hope any of you guys could help or just refer to some theorems!
$(f_n)$ converges to $f=0$
You have $$\Vert f_n - f\Vert_1 =\int_0^1 \vert f_n(x)\vert \ dx= \int_0^{1/n} (1-nx) \ dx =\left[x -\frac{nx^2}{2}\right]_0^{1/n}=\frac{1}{2n}$$ As $\lim\limits_{n \to \infty} \frac{1}{2n}=0$, we get our conclusion $f_n \to 0$.
$(g_n)$ doesn’t converge for the $\Vert \cdot \Vert_1$ norm
Suppose that $(g_n)$ converges to $g$ for the $\Vert \cdot \Vert_1$ norm. Let’s prove that this implies $g=0$. For all $n \in \mathbb N$, we have
$$\begin{aligned} \Vert g_n-g\Vert_1 &= \int_0^1 \vert g_n(x)-g(x)\vert \ dx\\ &=\int_0^{1/n} \vert g_n(x)-g(x)\vert \ dx + \int_{1/n}^1 \vert g_n(x)-g(x)\vert \ dx\\ &=\int_0^{1/n} \vert g_n(x)-g(x)\vert \ dx + \int_{1/n}^1 \vert g(x)\vert \ dx\\ &\ge \int_{1/n}^1 \vert g(x)\vert \ dx \end{aligned}$$
Moving $n \to \infty$ we get $0 \ge \int_0^1 \vert g(x) \vert \ dx$. As $g$ is supposed to be a continuous function, we get $g=0$. But this in contradiction with $\Vert g_n \Vert =1/2$ for all $n \in \mathbb N$.