I have a question regarding functional analysis:
Assume that $X$ is a strictly convex normed linear space and that $V ⊂ X$ is a proper
linear subspace. In addition, assume that for some element $x ∈ X$ a best approximation $v ∈ V$ of $x$ exists. Show that $v$ is unique.
I tried the following:
Suppose $v,v' \in V$ are best approximations of $x \in X$. Then,
$||x-v|| = ||x-v'||$ and the fact that $||x-v'||^2 =||x-v||^2+||v-v'||^2$ we have that $||v-v'|| = 0$. That is, if a best approximation $v \in V$ of $x$ exists then its unique.
Would this be a good approach? I have the feeling that I am missing things. Any help would be grateful. Thanks in advance.
I have read the following definition:
For each $x ∈ X$ there exists precisely one
best approximation in $V$ , namely there is a unique element $v ∈ V$ such that
$||x − v|| = d(x, V ) = inf{{||x − z|| : z ∈ V }}$.
I am not sure how to use this definition.
Best Answer
Suppose $\|x-v\|=\|x-v'\|$ with $v,v'(v\neq v')\in X$. Since $\frac {v+v'} 2 \in X$ we get $\|x-v\|=\|x-v'\| \leq \|x-\frac {v+v'} 2\|$. This gives $\|x-v\|+\|x-v'\|=\|(x-v)+(x-v')\|$. This implies that $x-v=c(x-v')$ for some scaler $c$. [See link below for definition of strict convexity].
But then $x \in X$ in which case we must have $v=v'=x$.
https://en.wikipedia.org/wiki/Strictly_convex_space