L2 norm satisfies triangle inequality on set of continuous functions $C[r_1,r_2]$
$begingroup$
Let $||f||_2 = (int_{r-1}^{r_2}|f|^2)^{1/2} $.
I have that
$star int_{r_1}^{r_2} |f cdot g| leq (int_{r_1}^{r_2} |f|^2)^{1/2} cdot (int_{r_1}^{r_2} |g|^2)^{1/2}$
for continuous functions $f,g$ on real interval $[r_1,r_2]$, and I want to show that for continuous functions $f,g,h$ on $[r_1,r_2]$, we have $||f-h||_2 leq ||f-g||_2 + ||g-h||_2$.
I have tried just proving this outright by looking at $||f-h||_2, ||f-g||_2, $ and $||g-h||_2$, but I can't see how $star$ relates to this problem. I'd just appreciate some hints as to how this proof works. Thanks!
analysis
asked Dec 7 '18 at 15:53
math177618math177618
816
$endgroup$
add a comment |
$begingroup$
Let $||f||_2 = (int_{r-1}^{r_2}|f|^2)^{1/2} $.
I have that
$star int_{r_1}^{r_2} |f cdot g| leq (int_{r_1}^{r_2} |f|^2)^{1/2} cdot (int_{r_1}^{r_2} |g|^2)^{1/2}$
for continuous functions $f,g$ on real interval $[r_1,r_2]$, and I want to show that for continuous functions $f,g,h$ on $[r_1,r_2]$, we have $||f-h||_2 leq ||f-g||_2 + ||g-h||_2$.
I have tried just proving this outright by looking at $||f-h||_2, ||f-g||_2, $ and $||g-h||_2$, but I can't see how $star$ relates to this problem. I'd just appreciate some hints as to how this proof works. Thanks!
analysis
asked Dec 7 '18 at 15:53
math177618math177618
816
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
1
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
1
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00
add a comment |
$begingroup$
Let $||f||_2 = (int_{r-1}^{r_2}|f|^2)^{1/2} $.
I have that
$star int_{r_1}^{r_2} |f cdot g| leq (int_{r_1}^{r_2} |f|^2)^{1/2} cdot (int_{r_1}^{r_2} |g|^2)^{1/2}$
for continuous functions $f,g$ on real interval $[r_1,r_2]$, and I want to show that for continuous functions $f,g,h$ on $[r_1,r_2]$, we have $||f-h||_2 leq ||f-g||_2 + ||g-h||_2$.
I have tried just proving this outright by looking at $||f-h||_2, ||f-g||_2, $ and $||g-h||_2$, but I can't see how $star$ relates to this problem. I'd just appreciate some hints as to how this proof works. Thanks!
analysis
asked Dec 7 '18 at 15:53
math177618math177618
816
$endgroup$
Let $||f||_2 = (int_{r-1}^{r_2}|f|^2)^{1/2} $.
I have that
$star int_{r_1}^{r_2} |f cdot g| leq (int_{r_1}^{r_2} |f|^2)^{1/2} cdot (int_{r_1}^{r_2} |g|^2)^{1/2}$
for continuous functions $f,g$ on real interval $[r_1,r_2]$, and I want to show that for continuous functions $f,g,h$ on $[r_1,r_2]$, we have $||f-h||_2 leq ||f-g||_2 + ||g-h||_2$.
I have tried just proving this outright by looking at $||f-h||_2, ||f-g||_2, $ and $||g-h||_2$, but I can't see how $star$ relates to this problem. I'd just appreciate some hints as to how this proof works. Thanks!
analysis
analysis
asked Dec 7 '18 at 15:53
math177618math177618
816
asked Dec 7 '18 at 15:53
math177618math177618
816
asked Dec 7 '18 at 15:53
math177618math177618
816
asked Dec 7 '18 at 15:53
math177618math177618
816
asked Dec 7 '18 at 15:53
math177618math177618
816
816
1
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
1
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
1
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
1
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
1
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00
1
1
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
1
1
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
1
1
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00
add a comment |
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1
$begingroup$
en.wikipedia.org/wiki/Minkowski_inequality#Proof
$endgroup$
– Federico
Dec 7 '18 at 15:56
1
$begingroup$
There are plenty of ways to go from Holder to Minkowski. This is just one of them
$endgroup$
– Federico
Dec 7 '18 at 15:59
1
$begingroup$
Have a look also here: math.stackexchange.com/questions/47930/…
$endgroup$
– Federico
Dec 7 '18 at 16:00