Recall that any finite dimensional normed linear space is complete. The following definition includes the
case where the norm comes from an inner product.
Definition 13.1.1Let
(H,(⋅,⋅))
be a complete inner product space. This means the norm comesfrom an inner product as described on Page 729,
|v|
≡
(v,v)
^{1∕2}. Such a space is called a Hilbertspace
As shown earlier, if H is finite dimensional, then it is a Hilbert space automatically. The following is the
definition of a convex set. This is a set with the property that the line segment between any two points in
the set is in the set.
Definition 13.1.2A nonempty subset K of a vector space is said to be convex if whenever x,y ∈ Kand t ∈
[0,1]
, it follows that tx +
(1− t)
y ∈ K.
Theorem 13.1.3Let K be a closed and convex nonempty subset of a Hilbert space and let y ∈ H. Alsolet
λ ≡ inf{|x− y| : x ∈ K }
Then if
{xn}
⊆ K is a sequence such that lim_{n→∞}
|xn − y|
=
|x − y|
, then it follows that
{xn}
is aCauchy sequence and lim_{n→∞}x_{n} = x ∈ K with
|x − y|
= λ. Also if
|x − y|
= λ =
|ˆx − y|
, then
ˆx
= x.
Proof:Recall the parallelogram identity valid in any innner product space:
2 2 2 2
|x + y| +|x− y| = 2 |x| + 2|y|
First consider the claim about uniqueness. Letting x,
and as n,m →∞, the right side converges to 0 by definition. Hence
{xn}
is a Cauchy sequence as claimed.
By completeness, it converges to some x ∈ H. Since K is closed, it follows that x ∈ K. Then from the
triangle inequality,
lim |y− xn| = |y− x| = λ.■
n→∞
In the above theorem, denote by Py the vector x ∈ K closest to y. It turns out there is an easy way to
characterize Py. For a given z ∈ K, one can consider the function t →
|x+ t(z − x) − y|
^{2} for x ∈ K. By
properties of the inner product, this is
2 2 2
t → |x− y| + 2tRe (z − x,x− y)+ t |z − x|
according to whether Re
(z − x,x − y)
≥ 0. Thus elementary considerations yield the two possibilities
shown in the graph. Either this function is increasing on
[0,1]
or it is not. In the case Re
(z − x,x− y)
< 0
the graph shows that x≠Py because there is a positive value of t such that the function is less than
|x − y|
^{2} and in case Re
(z − x,x− y)
≥ 0, we obtain x = Py if this is always true for any z ∈ K. Note that
by convexity, x + t
(z − x)
∈ K for all t ∈
[0,1]
since it equals
(1 − t)
x + tz.
PICT
Theorem 13.1.4Let x ∈ K and y ∈ H. Then there exists a closest point of K to y denoted by Py. Thenx = Py if and only if
2 2 2 2
|x+ t(z − x )− y|= |x − y| + 2tRe (z − x,x− y)+ t |z − x|
and is an increasing function on
[0,1]
. Thus it has its minimum at t = 0. In particular
|x − y|
^{2}≤
|z − y|
^{2}.
Since z is arbitrary, this shows x = Py.
Next suppose x = Py. Then for arbitrary z ∈ K, the minimum of t →
|x + t(z − x)− y|
^{2} occurs when
t = 0 since x + t
(z − x)
∈ K. This will not happen unless
Re (z − x,x − y) ≥ 0
because if this is less than 0, the minimum of that function will take place for some positive t. Thus 13.1
holds. ■
Every subspace is a closed and convex set. Note that Re and Im are real linear maps from ℂ to
ℝ.
Re (x+ iy) ≡ x, Im (x + iy) ≡ y
That is, for a,b real scalars and z,w complex numbers,
Re (az + bw) = aRe (z)+ bRe (w )
Im (az + bw) = aIm (z)+ bIm (w )
This assertions follow directly from the definitions of complex arithmetic and will be used
without any mention whenever convenient. The next proposition will be very useful in what
follows.
Then a graph of this along with the graph of y = x^{2} is given below. In this graph, the dashed
graph is of y = x^{2} and the solid line is the graph of the above Fourier series approximation.
PICT
If we had taken the partial sum up to n much bigger, it would have been very hard to distinguish between
the graph of the partial sum of the Fourier series and the graph of the function it is approximating. This is
in contrast to approximation by Taylor series in which you only get approximation at a point of a function
and its derivatives. These are very close near the point of interest but typically fail to approximate the
function on the entire interval.