How does one calculate the norm $||f_n||_{\infty} = sup_{x \in [0,1]} |f_n(x)|$?

Question

Let

$f_n:[0,1] \to \mathbb{R}$

$f_n(x) = 4n^2x, x\in [0,1/2n]$

$f_n(x) = -4n^2x + 4n, x\in [1/2n, 1/n]$

$f_n(x) = 0, else$

How can I calculate the norms

$||f_n||_{\infty} = sup_{x \in [0,1]} |f_n(x)|$

$||f_n||_{L^1} = \int_{0}^{1}|f_n(x)|dx$

I know how to calculate the euclidian norm normally, i.e. $\sqrt{(x_1^2 +...+x_n^2)}$, but I don't get it in this context. Aren't both norms equivalent? And what would happen for $n \to \infty$?

Answer 1

The norm $\|f\|_\infty$ is basically the largest value of $|f(x)|$ (as long as $f$ is continuous, which it is here). Here $f(x)\ge0$. On $[0,1/2n]$ the largest value of $f(x)$ is at $x=1/2n$ where $f(x)=2n$. I reckon the largest value on $[1/2n,1/n]$ is also $2n$.

Putting $f$ into the formula for $\|f\|_1$ gives $$\|f\|_1=\int_0^{1/2n}4n^2x\,dx+\int_{1/2n}^{1/n}(4n-4n^2x)\,dx$$ which I'm sure you can do. Alternatively, just look at the graph and write down the answer.