A linear operator on a finite dimensional Hilbert space is continuous

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?

Solution 1:

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.

cheers

math

Solution 2:

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

Solution 3:

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 $C = M.R$ .

Solution 4:

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.