What is the relation between a Banach space and a Hilbert space?

I know that a Hilbert space has inner product (and so a norm), but a Banach space has a just norm.

  • $\begingroup$ Have a look at this answer: qr.ae/TWI4tc. $\endgroup$
    – user168764
    May 4, 2019 at 23:03
  • 2
    $\begingroup$ This is a bit vague, but there are useful remarks that could be made in answers to it. Specifically, Hilbert spaces have a (true!) minimum/Dirichlet principle, and Banach spaces easily and non-pathologically fail this (e.g., the literally incorrect Dirichlet principle that was very important throughout the late 19th century, and only reformulated in 1905 by Giuseppe ("Beppo") Levi in 1905, and also Frobenius. (In Banach spaces, a non-empty closed convex subset need not have an element of least norm, and it could have infinitely-many. In Hilbert spaces there is a unique minimizing element.) $\endgroup$ May 4, 2019 at 23:09
  • 1
    $\begingroup$ @paulgarrett: Your (interesting!) comment draws my attention to this question. I have voted to reopen the post: the post provides enough context for anyone knows the basics of functional analysis (and it was posted on 2013 !). What a pity it would be if this post was deleted before your comment. I have not seen such comment/answer from an expert in this site for a very long time. Your words remind me of my old good times here. THANK YOU! $\endgroup$
    – user9464
    May 23, 2019 at 3:15

4 Answers 4


Hilbert spaces are a stricter subset of Banach spaces but they have the additional structure of an inner product which allows you to talk about orthonormal bases, unitary operators and so on. Example: Fourier transform theory is really beautiful on $L^2(\mathbb{R})$ but it is much more complicated on other Banach spaces because you don't have a notion of self-duality like in $L^2(\mathbb{R})$. You actually have to abstract a lot to define the Fourier transform on other $L^p$ spaces.

  • 2
    $\begingroup$ I think we should also mention the existence and uniqueness of best approximations in closed subspaces. This has far reaching consequences in other fields of science, such as physics and engineering (and, naturally, is very instrumental in Fourier analysis as well). $\endgroup$ Aug 20, 2013 at 16:54
  • 1
    $\begingroup$ Apart from Fourier Analysis, we may also consider the results of Approximation Theory. For the same reason Hilbert Space is much useful than Banach Space. $\endgroup$
    – Supriyo
    Aug 20, 2013 at 17:02

Hilbert spaces have an easier structure and are in a way (most often infinite dimensional) Euclidian spaces. However, many spaces of interest that are Banach spaces are not Hilbert spaces, hence they are important too.

To see if a Banach space is a Hilbert space it suffice to show that the norm satisfies the parallelogram law. In other words, if we have a Banach space $X$ such that $$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$$ for all $x,y\in X$ then $X$ is actually a Hilbert space.

To deduce how the norm looks like is a good exercise. In the real case (the complex case is similar) think of the expansions of $$\|x+y\|^2=\langle x+y,x+y\rangle$$ and $$\|x-y\|^2=\langle x-y,x-y\rangle.$$ The result is $$\langle x,y\rangle=\frac{\|x+y\|^2-\|x-y\|^2}{4}.$$


I don't think they are "better", per se. A Hilbert space is a very special type of Banach space - one which is meant to generalize familiar notions from $\mathbb{R}^n$. (For instance, you can quite naturally speak of when two vectors in Hilbert space are orthogonal).

In general, Hilbert spaces are "easier" to understand than general Banach spaces, and are usually a good place to start if you are learning the subject (For instance, try to see why the Hahn-Banach theorem is much simpler for Hilbert spaces)


Since most of the time we deal with $\mathbb{R}^n$, you can say that one has advantage over the other in $\mathbb{R}^n$. So I consider $\mathbb{R}^n$ and explain why Hilbert spaces are more useful in practice:

Hilbert space is defined as a vector space $H$ equipped with an inner product $\langle \cdot,\cdot \rangle$, consequently this inner product induces a norm which we call it a metric, $d(\cdot,\cdot)$. Hence, it is the time to define a new vector space $M$ endowed with the mentioned metric $d(\cdot,\cdot)$ which is a metric vector space. Now if every Cauchy sequence in $M$ converges to a point in $M$ we have Banach space which is a complete metric space. In $\mathbb{R}^n$ the relation is as follows:

enter image description here

Keep in mind that in a vector space $M$, the only tool that we have is the metric $d(\cdot,\cdot)$, so we can talk about length of a vector or distance between vectors.

However, when we have inner product that satisfies Parallelogram law, we also have an inner product through which we can define projection from one vector to another which is of interest in practice because we can talk about the angles. Using this concept, optimization problems such as minimum distance to a subspace can be defined as projection of that point onto a set.


Not the answer you're looking for? Browse other questions tagged .