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

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 3 '18 at 16:22
• @masacroso True ! I just tried to finish it to have the standard bounded form. – Rebellos Nov 3 '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 4 '18 at 19:14

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.