# Thread: inequality with two metrics

1. ## inequality with two metrics

I want to prove that $sup \ x \in C[0,1] |f_n(x)-f(x) | \geq ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}$

$({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}$
$\leq$ (by c-s inequality)
$({\displaystyle\int^1_0 |f_n(x)|^2 \ dx})^{1/2} + ({\displaystyle\int^1_0 |f(x)|^2 \ dx})^{1/2}$
$\leq$
$sup |f_n(x)| + sup|f(x)|$

However I am unsure where to go now

2. ## Re: inequality with two metrics

Originally Posted by FGT12
I want to prove that $sup \ x \in C[0,1] |f_n(x)-f(x) | \geq ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}$

$({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}$
$\leq$ (by c-s inequality)
$({\displaystyle\int^1_0 |f_n(x)|^2 \ dx})^{1/2} + ({\displaystyle\int^1_0 |f(x)|^2 \ dx})^{1/2}$
$\leq$
$sup |f_n(x)| + sup|f(x)|$

However I am unsure where to go now
If $M = \sup_{x\in[0,1]}|f_n(x)-f(x)|$ then $|f_n(x)-f(x)|\leqslant M$ for all x in [0,1]. It follows that $\int_0^1|f_n(x)-f(x)|^2dx\leqslant\int_0^1M^2dx = M^2.$ Take the square root of both sides to get the inequality that you want (no need for Cauchy–Schwarz).