daniel_i_l
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1) C[-1,2] is a space of all continues functions f: [-1,2] -> C (complex)
Is:
<f,g> = \int_{-1}^{2}|f(t) + g(t)|dt
an inner product of C[-1,2]?
I think that the answer is no because:
<f+g, h> \neq <f,h> + <g,h>
for some f and g. this can happen when all the functions are positive and so:
|f(t) + h(t) + g(t)| doesn't equal |f(t) + h(t)| + |g(t) + h(t)|
2)
V is a space of all real functions with defined double derivatives in the interval
[-\pi, \pi]
we define:
<f,g> = f(-\pi)g(-\pi) + \int_{-\pi}^{\pi}f''(x)g''(x)dx
is <f,g> an inner product of V?
also here i think that the answer is no because <f,f> can equal zero even if all of f isn't 0, this can happen if f(-pi) is zero and the double derivative is 0 everywere (contiues slope).
am i correct?
Thanks.
Is:
<f,g> = \int_{-1}^{2}|f(t) + g(t)|dt
an inner product of C[-1,2]?
I think that the answer is no because:
<f+g, h> \neq <f,h> + <g,h>
for some f and g. this can happen when all the functions are positive and so:
|f(t) + h(t) + g(t)| doesn't equal |f(t) + h(t)| + |g(t) + h(t)|
2)
V is a space of all real functions with defined double derivatives in the interval
[-\pi, \pi]
we define:
<f,g> = f(-\pi)g(-\pi) + \int_{-\pi}^{\pi}f''(x)g''(x)dx
is <f,g> an inner product of V?
also here i think that the answer is no because <f,f> can equal zero even if all of f isn't 0, this can happen if f(-pi) is zero and the double derivative is 0 everywere (contiues slope).
am i correct?
Thanks.
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