How do I show that a linear function from a Hilbert space $H$ to itself is continuous if $H$ is finite dimensional?

Also, what would be an example of a linear function from a Hilbert space to itself which is not continuous?

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    $\begingroup$ Do you know that a linear operator between Banach spaces is continuous if and only if it is bounded? (You should do this as an exercise first if not.) Can you show that a linear operator from a finite-dimensional Hilbert space to another one must be bounded? Can you write down a linear operator on an infinite-dimensional Hilbert space which isn't bounded? $\endgroup$ – Qiaochu Yuan Sep 15 '11 at 3:59
  • $\begingroup$ An aside: An example of a discontinuous linear operator on a Hilbert space requires some choice: mathoverflow.net/questions/5303/basis-of-linfinity/5313#5313 $\endgroup$ – Jonas Meyer Sep 15 '11 at 4:06

A little bit more general:

Let $\ T \,\colon X \to Y\:$ be a linear operator between two normed vector spaces, where $X$ should be finite dimensional. Then every linear map is continuous.

Proof: Define the "graph norm" induced by $T$.

$\lVert x\rVert_T :=\lVert x\rVert_X + \lVert Tx\rVert_Y $. This is a norm. Now use the fact, that on a finite dimensional vector space every two norms are equivalent. So there must be a constant $\lambda $ such that $\lVert x\rVert_T \le \lambda \lVert x\rVert_X $.

Clearly the following inequality holds: $$\lVert Tx\rVert_Y \le \lVert x\rVert_T. $$

Combine this two fact, it's obvious that $ T $ is bounded and therefore, as Yuan Qiaochu said, $T$ must be continuous.

Hopefully I considered all rules and conventions in this forum, since this is my first answer. If not, I'm very sorry.



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    $\begingroup$ What a beautiful proof, math! Thanks! $\endgroup$ – user16185 Sep 16 '11 at 17:25
  • $\begingroup$ Nice proof! But I would remove reference to Hellinger - Toeplitz, I don't see how it fits here. Just my opinion, though. $\endgroup$ – Giuseppe Negro Sep 16 '11 at 17:36
  • $\begingroup$ @Davide Giraudo: Thx for editing my answer and the better proof-writing! $\endgroup$ – math Sep 17 '11 at 15:00
  • $\begingroup$ @math you're welcome. $\endgroup$ – Davide Giraudo Sep 17 '11 at 15:15

The intuitive way (as suggested by GEdgar):

Since $X$ is finite dimensional, there is a finite basis $ u_1, \ldots , u_n$ for $X$. Using the Gram-Schmidt orthonormalisation process, you can use this basis to construct an orthonormal basis $ v_1,\ldots , v_n$ for $X$.

Now, if $x$ is any unit vector in $X$, $x$ can be written as $\displaystyle \sum_{i=1}^n \lambda_i v_i $, for some $ \lambda_i \in \mathbb{R}$ satisfying $\displaystyle \sum_{i=1}^{n} \vert \lambda_i \vert = 1$.

But then, \begin{align*} \| Tx \|&= \lVert\sum_{i=1}^{n} \lambda_i \ T v_i \rVert \\ &\leq \sum_{i=1}^{n} | \lambda_i | \ \| T v_i \| \\ &\leq \left( \sum_{i=1}^{n} \lambda_i \right) \cdot \max_{1 \leq i \leq n} \| T v_i \| \\ &= \max_{1 \leq i \leq n} \| T v_i \| < + \infty. \end{align*}

From this inequality we see that indeed T must be bounded (i.e. contiunuous) with $\displaystyle \| T \| \leq \max_{1 \leq i \leq n} \| T v_i \| $ (in fact we must have equality since the $ v_i $ are also admissible unit vectors!)

Best regards,

Benno Handsma

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  • $\begingroup$ Nice! (Please see comments to GEdgard answer). $\endgroup$ – Giuseppe Negro Sep 18 '11 at 1:09
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    $\begingroup$ This argument is somewhat flawed: note that a unit vector can only be written as $\sum \lambda_i v_i$ with $\sum |\lambda_i|^2 = 1$ (note the squares). You can save it by simply putting the norm $v = \sum |\lambda_i|$ instead of the Hilbert norm and drop mention of Gram-Schmidt. @Giuseppe: ping! $\endgroup$ – t.b. Sep 18 '11 at 12:31

Another way: Choose a basis, then your linear transformation is given by a matrix. Check that the formulas for the transformation in terms of the matrix are continuous.

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    $\begingroup$ This is the most intuitive approach for sure. However, you still need to prove that the mapping $H \to \mathbb{R}^n$ arising from the choice of a basis is continuous. I'm afraid this is as difficult as proving the original claim. $\endgroup$ – Giuseppe Negro Sep 16 '11 at 17:34
  • $\begingroup$ @Giuseppe: I don't understand this objection when at the same time you don't object to applying the same fact in math's answer (the statement that all norms on a finite-dimensional space are equivalent is just another way of stating the same). $\endgroup$ – t.b. Sep 17 '11 at 13:37
  • $\begingroup$ @Theo: I've thought about your counter-objection a little bit. Reading your comment I realized that the following three facts about finite-dimensional normed spaces are equivalent, that is, once one is proven the other two follow easily: 1) Every linear operator is continuous; 2) For every choice of a basis the coordinate functions are continuous; 3) Every two norms are equivalent. In a general setting neither one of those statements is too easy to prove: they more or less follow from compactness of the identity operator, as far as I see. (continue...) $\endgroup$ – Giuseppe Negro Sep 18 '11 at 1:02
  • $\begingroup$ On the contrary, in Hilbertian setting we can take a shortcut ans prove 1) directly by means of the concept of orthonormal basis, as Benno shows us below. 2) and 3) are also easy if we require orthonormality in 2) and Hilbertian norms in 3). Once again we see that Hilbert spaces are "much nicer" than their uncivilized cousins, Banach spaces. :-) $\endgroup$ – Giuseppe Negro Sep 18 '11 at 1:08
  • $\begingroup$ @Giuseppe: I essentially agree with what you say in the first comment. Let me add the following observation (which I think is the crucial fact used in this whole thread): The spaces $\ell^1(S)$ have the following property (in fact, they are characterized by a version of it by a theorem of Pełczyński and Köthe): If $X$ is a Banach space and $f: S \to X$ is a function with $f(S)$ bounded then there exists a unique bounded linear operator $F: \ell^{1}(S) \to X$ such that $F|_S = f$ and $F$ has norm $\|F\| = \sup_{s \in S}\|f(s)\|$. If $S$ is finite, the boundedness condition on $f(S)$ is void. $\endgroup$ – t.b. Sep 18 '11 at 12:28

Here's another similar solution:

Let $\beta = \{e_1, e_2, \dots, e_n \}$ be a basis for $X$. Then for all $x$ in $X$ we have : $$\|Tx\| = \| \sum_{k=1}^n{\alpha_k Te_k}\| \le \sum_{k=1}^n{\| \alpha_kTe_k\|}=\sum_{k=1}^n{\vert \alpha_k \vert \ \| Te_k\|} $$

Let $\displaystyle \max_{1 \leq k \leq n} \| T e_k \| = M.$ Then $\|Tx\| \le M \sum_{k=1}^n{\vert \alpha_k \vert}$.

But $\| x\|_1 = \sum_{k=1}^n{\vert \alpha_k \vert}$ is a norm as you may check. So $\|Tx\| \le M \|x\|_1$ and since all norms on a finite dimensional space are equivalent there exists $R >0$ such that $\|x\|_1 \le R\|x\|$ and thus $$\|Tx\| \le C \|x\|$$ where $C = M.R$ .

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