Every linear operator $T:X \to Y$ on a finite-dimensional normed space is bounded

Question

Exercise :

Show that if $X$ is a finite-dimensional normed space and $Y$ is a normed space, then every linear operator $T:X \to Y$ is bounded.

Attempt :

Since $X$ is finite-dimensional, say $\dim(X)=n$, there exists a basis $\{e1,e2,...,en\}$ of $X$ such that every element $x\in X$ can be written uniquely in the form:

$$\begin{align} \quad x = a_1e_1 + a_2e_2 + ... + a_ne_n \end{align}$$

where $a_1,a_2,\dots,a_n \in \mathbb R$.

Now, $\forall \; x \in X$, it is :

$$\begin{align} \quad \| T(x) \|_Y &= \| T (a_1e_1 + a_2e_2 + ... + a_ne_n) \|_Y \\ &= \| a_1 T(e_1) + a_2 T(e_2) + ... + a_n T(e_n) \|_Y \\ & \leq \sum_{k=1}^{n} |a_k| \| T(e_k) \|_Y \end{align}$$

Using the Cauchy-Schwarz inequality, we yield :

$$\begin{align} \quad \| T(x) \| & \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \left ( \sum_{k=1}^{n} \| T(e_k) \|_Y^2 \right )^{1/2} \\ & \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} \cdot M \end{align}$$

But regarding equivalence of norm in correlation to finite-dimensional spaces, we have that :

$$\begin{align} \quad \| T(x) \|_Y & \leq M \| x \|_* \end{align}$$

Then, $\exists c_1,c_2 \in \mathbb R^+ : \forall x \in X$ it is :

$$\begin{align} \quad c_1 \| x \|_X \leq \| x \|_* \leq c_2 \| x \|_X \end{align}$$

Thus $\forall x \in X$ it is :

$$\begin{align} \quad \| T(x) \| & \leq c_1M \| x \|_X \end{align}$$

which tells us that $T$ is bounded.

Question : It seemed like a rather hard exercise to me so I am not sure if my proof/approach is definitely correct or rigorous enough. Any insight will be very helpful !

from the inequality $| T(x) |Y \leq \sum{k=1}^{n} |a_k| | T(e_k) |_Y$ the proof is almost done because clearly $|T(x)|_Y<\infty$, so you have a linear operator that map bounded sets to bounded sets, what is the definition of bounded linear operator. — Masacroso, Nov 03 '18 at 16:22
@masacroso True ! I just tried to finish it to have the standard bounded form. — Rebellos, Nov 03 '18 at 16:32
@Masacroso: "bounded" is not a magical word. The notion of "bounded" depends on a metric. It is not obvious that different metrics will give you the same bounded sets. It does work for metrics given by norms on a finite-dimensional space, because of the nontrivial fact that all norms are equivalent, as used by the OP. — Martin Argerami, Nov 04 '18 at 19:14

score 12 · Accepted Answer · answered Nov 04 '18 at 19:16

12

Your proof is both correct and rigorous. The only change I would suggest is that, instead of using Cauchy-Schwarz, you can take $M=\max\{\|Te_k\|:\ k=1,\ldots,n\}$ and then $$ \sum_{k=1}^{n} |a_k| \| T(e_k) \|_Y \leq M\,\sum_{k=1}^n|a_k|, $$ and you can use that norm to compare with the original.

answered Nov 04 '18 at 19:16

Martin Argerami

205,756

Thanks a lot for your input ! This is a pretty nice suggestion and it simplifies the proof a bit (makes it a lot shorter). Would it be more "elegant" to approach it like that ? – Rebellos Nov 04 '18 at 21:42
There's no canonical answer. But I'm all for typing less. – Martin Argerami Nov 04 '18 at 22:15
@MartinArgerami how can one compare $\displaystyle\sum_{k = 1}^{n} |a_k|$ to $|a|$ (I see that the former is larger, but that doesn't help us with the inequality)? – Rhi Jun 20 '21 at 14:29
4

@Rhi: that requires the well-known argument that all norms are equivalent in a finite-dimensional space. – Martin Argerami Jun 20 '21 at 19:59

score 2 · Answer 2 · answered Dec 08 '21 at 19:55

2

I think you must use the suggestion by @MartinArgerami because you do not specify that $Y$ is an inner product space so there is no Cauchy-Schwarz inequality in $Y$.

If $Y$ is an inner product space, I don't recognize the inequality you used as the Cauchy-Schwarz inequality.

answered Dec 08 '21 at 19:55

tjevans

75

1

But the inequality was applied for reals numbers, not in Y. – BlueNight Feb 03 '23 at 02:01

Every linear operator $T:X \to Y$ on a finite-dimensional normed space is bounded

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