QUIZ 4 (ANALYSIS II)
Feb 14, 2020 (in tutorial) Name: Solutions
In what follows,K∈ {R,C}. The symbolδij is the Kronecker symbol and feel free to ask the TA’s what it means if you don’t know. Finally,R+= [0,∞).
(1) Let (V,k k) be a finite dimensional normed linear space over K and W a vector subspace ofV. Show thatW is closed inV.
Solutions: We knowW ∼=Kd for some dand that all norms on Kd are equivalent, whence Kd is complete in all norms on it, since it is so with respect tok k∞. It follows thatW is complete in V. It is immediate that W is closed. Indeed, ifw∗ is in the closure ofW, we have a sequence{wn} inW which converges tow∗, and sinceW is complete,w∗ must lie inW.
(2) Leth, ibe an inner product on a finite dimensionalK-vector spaceV and letk k: V → R+ be the usual norm, namely kvk = hv, vi12. Let W be a subspace of V and w1, . . . , wk a basis of W such that hwi, wji = δij for i, j ∈ {1, . . . , k}. Let π: V →W be the map π(v) = Pk
i=1hv, wiiwi. Fix v ∈ V and define fv: W → R+ by the formula fv(w) = kw−vk. Show that
(a) fv is continuous onW;
Solutions: This follows from the inequality |kw−vk − kw0−vk| ≤ kw−w0k. It is immediate that if >0 is given, pickingδ= we get
|fv(w)−fv(w0)|< wheneverkw−w0k< δ.
(b) fv attains its minimum at π(v) and if fv(w) = fv(π(v)) for some w∈W, thenw=π(v).
Solutions: Ifw∈W then we claim that hv−π(v), wi= 0.
Indeed
hv−π(v), wji=hv, wji −
k
X
i=1
hhv, wiiwi, wji
=hv, wji − hv, wji= 0.
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In particularhv−π(v), π(v)i= 0. Forw∈W, we therefore have (by an easy calculation):
fv(w)2=hv−w, v−wi
=hv−πv, v−π(v)i+hπ(v)−w, π(v)−wi
=fv(π(v))2+kπ(v)−wk2
It follows that fv(π(v)) ≤ fv(w) for w ∈ W, and when w = π(v) this minimum is attained. Finally, iffv(w) =fv(π(v)) then the above shows thathπ(v)−w, π(v)−wi= 0, whencew=π(v).
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