Functions not satisfying parallelogram identity with supremum norm

JackTheLad
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Homework Statement


Find two functions f, g \in C[0,1] (i.e. continuous functions on [0,1]) which do not satisfy

2 ||f||^2_{sup} + 2 ||g||^2_{sup} = ||f+g||^2_{sup} + ||f-g||^2_{sup}

(where || \cdot ||_{sup} is the supremum or infinity norm)

Homework Equations


Parallelogram identity: 2||x||^2 + 2||y||^2 = ||x+y||^2 + ||x-y||^2 holds for any x,y


The Attempt at a Solution


Honestly no idea.
 
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Just try some functions. It's really not hard to find an example that doesn't work.
 
For posterity, two functions which fit nicely are
f(x) = x
g(x) = x-1


(I had tried lots of functions but they worked; not very helpful response)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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