Spaces that narrowly miss becoming Hilbert Spaces

A Hilbert space is a vector space with:

1. an inner product, which defines lengths and angles; and
2. completeness—every Cauchy sequence converges to an element still inside the space.

Several important spaces miss by just one condition.

1. Polynomials with the $L^2$ inner product: not complete

Let$$P[0,1]=\{\text{all polynomials on }[0,1]\}$$

with$$\langle p,q\rangle=\int_0^1 p(x)\overline{q(x)}\,dx.$$

This is a valid inner-product space. But a sequence of polynomials can converge in $L^2$ to a non-polynomial function, such as $e^x$.

Because the limit is outside $P[0,1]$, the space is not complete and therefore not Hilbert.

Its completion is $L^2[0,1]$, which is a Hilbert space.

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2. Finite sequences: an “unfinished” $\ell^2$

Consider$$c_{00}=\{(x_1,x_2,\ldots):\text{only finitely many }x_n\neq0\}$$

with the usual inner product$$\langle x,y\rangle=\sum_{n=1}^{\infty}x_n\overline{y_n}.$$

Every vector has finite support, so the sum is well defined. Now consider$$x^{(N)}=\left(1,\frac12,\frac13,\ldots,\frac1N,0,0,\ldots\right).$$

This is a Cauchy sequence, but its limit would be$$\left(1,\frac12,\frac13,\ldots\right),$$

which has infinitely many nonzero entries and therefore is not in $c_{00}$.

So $c_{00}$ is an inner-product space but not complete. Its completion is $\ell^2$.

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3. Continuous functions with the $L^2$ norm: limits can become discontinuous

Take $C[0,1]$, the continuous functions on $[0,1]$, with$$\langle f,g\rangle=\int_0^1 f(x)\overline{g(x)}\,dx.$$

Continuous functions can converge in this norm to a discontinuous function—for example, increasingly sharp continuous approximations to a step function.

Thus $C[0,1]$ is not complete under the $L^2$ norm. Once again, its completion is $L^2[0,1]$.

A subtle point: $C[0,1]$ is complete under the supremum norm, but that norm does not make it a Hilbert space.

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4. $C[0,1]$ with the supremum norm: complete, but no compatible inner product

Define$$\|f\|_\infty=\max_{x\in[0,1]}|f(x)|.$$

This makes $C[0,1]$ a complete normed space—a Banach space. However, the norm does not come from an inner product.

A norm induced by an inner product must satisfy the parallelogram identity:$$\|f+g\|^2+\|f-g\|^2 = 2\|f\|^2+2\|g\|^2.$$

The supremum norm does not always satisfy this identity. Therefore, $C[0,1]$ with $\|\cdot\|_\infty$ is Banach but not Hilbert.

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5. $\ell^p$ for $p\neq2$: complete, but with the wrong geometry

For $1\le p<\infty$,$$\ell^p=\left\{x:\sum_{n=1}^{\infty}|x_n|^p<\infty\right\}, \qquad \|x\|_p=\left(\sum |x_n|^p\right)^{1/p}.$$

These spaces are complete. But when $p\neq2$, their usual norm does not come from an inner product.

For example, take$$x=(1,0),\qquad y=(0,1).$$

In $\ell^p$,$$\|x+y\|_p=\|x-y\|_p=2^{1/p}.$$

The parallelogram identity would require$$2\cdot2^{2/p}=4,$$

which holds only when $p=2$.

Thus $\ell^1,\ell^3,\ell^4,\ldots$ are Banach spaces but not Hilbert spaces.

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6. $L^p[0,1]$ for $p\neq2$: complete, but not inner-product spaces

Similarly,$$\|f\|_p=\left(\int_0^1|f(x)|^p\,dx\right)^{1/p}$$

makes $L^p[0,1]$ complete for $p\ge1$. But its usual norm comes from an inner product only for $p=2$.

So $L^1$ and $L^3$, for example, have well-behaved limits but lack Hilbert-space notions such as orthogonality, angles, and orthogonal projection.

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7. Minkowski spacetime: has a bilinear form, but it is not positive definite

Special relativity uses$$\langle x,x\rangle_M = -x_0^2+x_1^2+x_2^2+x_3^2.$$

This resembles an inner product, but it can be negative or zero even for a nonzero vector. A lightlike vector satisfies$$\langle x,x\rangle_M=0 \quad\text{while}\quad x\neq0.$$

It therefore fails the positive-definiteness requirement. Minkowski space is an indefinite inner-product space, not a Hilbert space.

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8. Smooth functions: too restrictive to contain all their limits

Consider $C^\infty[0,1]$, the infinitely differentiable functions, with the $L^2$ inner product.

A sequence of smooth functions may converge in $L^2$ to a nonsmooth or discontinuous function. The limit therefore leaves $C^\infty[0,1]$.

So the space possesses an inner product but is not complete in the resulting norm.

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The two most common “near misses” are therefore:

Space	Inner product?	Complete?	Hilbert?
$c_{00}$ with $\ell^2$ norm	Yes	No	No
Polynomials with $L^2$ norm	Yes	No	No
$C[0,1]$ with $L^2$ norm	Yes	No	No
$C[0,1]$ with sup norm	No compatible one	Yes	No
$\ell^p,\ p\neq2$	No compatible one	Yes	No
$L^p,\ p\neq2$	No compatible one	Yes	No
Minkowski space	Indefinite form	Finite-dimensional	No
$\ell^2$ or $L^2$	Yes	Yes	Yes

The closest misses are $c_{00}$, polynomials, and smooth functions: they already possess the correct Hilbert-space geometry but contain “holes.” Completing them fills those holes and produces a genuine Hilbert space.