Triangle inequality in bracket space proof
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Consider ∥u+v∥2=(u+v)⋅(u+v)‖u+v‖2=(u+v)⋅(u+v)where u⋅vu⋅v represents the standard inner product/scalar product.Therefore
∥u+v∥2=∥u∥2+2(u⋅v)+∥v∥2.‖u+v‖2=‖u‖2+2(u⋅v)+‖v‖2.
By the Cauchy-Schwarz Inequality we have
u⋅v≤∥u∥⋅∥v∥.u⋅v≤‖u‖⋅‖v‖.
So,
∥u+v∥2=∥u∥2+2(u⋅v)+∥v∥2≤∥u∥2+2∥u∥⋅∥v∥+∥v∥2=(∥u∥+∥v∥)2,‖u+v‖2=‖u‖2+2(u⋅v)+‖v‖2≤‖u‖2+2‖u‖⋅‖v‖+‖v‖2=(‖u‖+‖v‖)2,
i.e.,
∥u+v∥2≤(∥u∥+∥v∥)2⟹∥u+v∥≤∥u∥+∥v∥.‖u+v‖2≤(‖u‖+‖v‖)2⟹‖u+v‖≤‖u‖+‖v‖.
The Cauchy-Schwarz Inequality holds for any inner Product, so the triangle inequality holds irrespective of how you define the norm of the vector to be, i.e., the way you define scalar product in that vector space.
In this case, the equality holds when vectors are parallel i.e, u=kvu=kv, k∈R+k∈R+because u⋅v=∥u∥⋅∥v∥cosθu⋅v=‖u‖⋅‖v‖cosθ when cosθ=1cosθ=1, the equality of the Cauchy-Schwarz inequality holds.
∥u+v∥2=∥u∥2+2(u⋅v)+∥v∥2.‖u+v‖2=‖u‖2+2(u⋅v)+‖v‖2.
By the Cauchy-Schwarz Inequality we have
u⋅v≤∥u∥⋅∥v∥.u⋅v≤‖u‖⋅‖v‖.
So,
∥u+v∥2=∥u∥2+2(u⋅v)+∥v∥2≤∥u∥2+2∥u∥⋅∥v∥+∥v∥2=(∥u∥+∥v∥)2,‖u+v‖2=‖u‖2+2(u⋅v)+‖v‖2≤‖u‖2+2‖u‖⋅‖v‖+‖v‖2=(‖u‖+‖v‖)2,
i.e.,
∥u+v∥2≤(∥u∥+∥v∥)2⟹∥u+v∥≤∥u∥+∥v∥.‖u+v‖2≤(‖u‖+‖v‖)2⟹‖u+v‖≤‖u‖+‖v‖.
The Cauchy-Schwarz Inequality holds for any inner Product, so the triangle inequality holds irrespective of how you define the norm of the vector to be, i.e., the way you define scalar product in that vector space.
In this case, the equality holds when vectors are parallel i.e, u=kvu=kv, k∈R+k∈R+because u⋅v=∥u∥⋅∥v∥cosθu⋅v=‖u‖⋅‖v‖cosθ when cosθ=1cosθ=1, the equality of the Cauchy-Schwarz inequality holds.
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