Question #150758
Show that (C[a, b], ||f||2) is a norm
1
Expert's answer
2020-12-15T13:37:43-0500

First of all, for fC([a;b])f\in\mathcal{C}([a;b]) we see that

f2=(abf(x)2dx)1/2||f||_2 = (\int_a^b |f(x)|^2 dx)^{1/2}

is well defined and is in [0;+)[0;+\infty) .

Now let's verify the norm properties:

  1. Separation, f2=0f=0||f||_2 = 0 \leftrightarrow f=0 . If f=0,f2=0f=0, ||f||_2=0 is obvious. Now suppose f0,f20f\neq 0, |f|^2 \neq 0 is a continuous positive function and thus abf(x)2dx0\int_a^b |f(x)|^2 dx \neq 0, therefore f20||f||_2\neq 0 .
  2. Linearity, αf2=αf2||\alpha f||_2 = |\alpha| \cdot||f||_2 . It follows from linearity of an integral : (abαf2(x)dx)1/2=(α2abf(x)2dx)1/2=αf2(\int_a^b |\alpha f|^2(x) dx)^{1/2} = (|\alpha|^2 \int_a^b |f(x)|^2 dx)^{1/2}=|\alpha|\cdot||f||_2
  3. Triangle inequality, f+g2f2+g2||f+g||_2 \leq ||f||_2 + ||g||_2 .

abf+g2(x)dxabf2(x)dx+abg2(x)dx+2abfg(x)dx\int_a^b|f+g|^2(x) dx \leq \int_a^b|f|^2(x) dx +\int_a^b|g|^2(x)dx + 2 \int_a^b |fg| (x) dx

We have (abfg(x)dx)2(abf(x)2dx)(abg(x)2dx)(\int_a^b |fg|(x) dx)^2 \leq (\int_a^b |f(x)|^2 dx) \cdot (\int_a^b |g(x)|^2 dx), as

fg(abf(x)2dx)1/2(abg(x)2dx)1/2f2abf(x)2dx+g2abg(x)2dx2\frac{|fg|}{(\int_a^b |f(x)|^2 dx)^{1/2}\cdot(\int_a^b |g(x)|^2 dx)^{1/2}} \leq \frac{\frac{|f|^2}{\int_a^b |f(x)|^2 dx}+\frac{|g|^2}{\int_a^b |g(x)|^2 dx}}{2} (by applying 2aba2+b22|ab|\leq a^2+b^2 to f(abf(x)2dx)1/2\frac{|f|}{(\int_a^b |f(x)|^2 dx)^{1/2}} and g(abg(x)2dx)1/2\frac{|g|}{(\int_a^b |g(x)|^2 dx)^{1/2}} ) and thus by integrating on [a;b][a;b] we find :

abfg(x)dx(abf(x)2dx)1/2(abg(x)2dx)1/21\frac{\int_a^b|fg| (x)dx}{(\int_a^b |f(x)|^2dx)^{1/2}\cdot(\int_a^b |g(x)|^2 dx)^{1/2}} \leq 1

Therefore (abfg(x)dx)2(abf2dx)(abg2dx)(\int_a^b |fg|(x) dx)^2 \leq (\int_a^b |f|^2 dx) \cdot (\int_a^b |g|^2 dx) .

And we conclude :

f+g22=abf+g2abf2+abg2+2abfg||f+g||_2^2 = \int_a^b|f+g|^2 \leq \int_a^b|f|^2 +\int_a^b |g|^2 +2\int_a^b |fg|

f+g22abf2+abg2+2(abf2)1/2(abg2)1/2||f+g||_2^2 \leq \int_a^b|f|^2 +\int_a^b|g|^2 +2(\int_a^b|f|^2)^{1/2}(\int_a^b|g|^2)^{1/2}

f+g22(f2+g2)2||f+g||_2^2 \leq (||f||_2+||g||_2)^2

And thus f+g2f2+g2||f+g||_2 \leq ||f||_2 + ||g||_2 .




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