- #1
D_Miller
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I am taking a self-study individual course on convex analysis but I'm having some troubles with the basics as I'm trying to do the exercises in my notes.
I'm asked to consider the space [tex]C[0,1][/tex] of continuous, complex-valued functions on [tex][0,1][/tex], equipped with the supremum norm [tex]\|\cdot\|_{\infty}[/tex]. Let [tex]M[/tex] be the subset of [tex]C[0,1][/tex] consisting of all those functions [tex]f[/tex] such that
[tex]\int_{0}^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^{1}f(x)dx=1[/tex].
I now have to show that [tex]M[/tex] is convex and closed in [tex]C[0,1][/tex]. This is vastly different from anything else I've done in this course so far, and I really have no clue how to start. Any help would be greatly appreciated.
I'm asked to consider the space [tex]C[0,1][/tex] of continuous, complex-valued functions on [tex][0,1][/tex], equipped with the supremum norm [tex]\|\cdot\|_{\infty}[/tex]. Let [tex]M[/tex] be the subset of [tex]C[0,1][/tex] consisting of all those functions [tex]f[/tex] such that
[tex]\int_{0}^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^{1}f(x)dx=1[/tex].
I now have to show that [tex]M[/tex] is convex and closed in [tex]C[0,1][/tex]. This is vastly different from anything else I've done in this course so far, and I really have no clue how to start. Any help would be greatly appreciated.
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