# inequality with two metrics

• Nov 16th 2011, 08:51 AM
FGT12
inequality with two metrics
I want to prove that $\displaystyle sup \ x \in C[0,1] |f_n(x)-f(x) | \geq ({\displaystyle\int^1_0 |f_n(x)-f(x)|^2 \ dx})^{1/2}$

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

However I am unsure where to go now
• Nov 16th 2011, 11:26 AM
Opalg
Re: inequality with two metrics
Quote:

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

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

However I am unsure where to go now

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