# Prove a matrix function is continuous

Suppose $$f:(C[0,2],\lVert\rVert_{\infty})\to (M_{2}(\mathbb{R}), \lVert T\rVert_{op})$$, where $$\lVert T\rVert_{op} = \sup_{\lVert \mathbf{x}\rVert =1}\lVert T\mathbf{x}\rVert$$

Let $$A=U^{-1}DU$$ be some diagonalisable matrix, where $$D$$ is a diagonal matrix with entries in $$[0,2]$$ and define $$f(g)=g(A)=U^{-1}g(D)U$$. (Note: $$g(D)$$ means apply the function $$g$$ to each entry of the matrix $$D$$). Prove that the function $$f$$ is continuous.

It seems I need need to use the epsilon-delta definition, but I do not know how to simplify the expression $$\lVert f(g)-f(h)\rVert_{op}=\sup_{\lVert \mathbf{x}\rVert=1}\lVert U^{-1}[g(D)-h(D)]U\mathbf{x}\rVert$$ in order to relate it to $$\lVert g-h\rVert_{\infty}$$. How can I do this?

• If I correctly understood your question. You need to show that for any $\epsilon>0$ you can find $\delta>0$ so that for any two functions $h, g$ such that $\|h - g\|_\infty <\delta$ you have $\|f(h) - f(g)\|_{op} < \varepsilon$ – them Aug 2 '19 at 14:26
• What is $\lVert f(h)-f(g)\rVert$ explicitly? – BaroqueFreak Aug 2 '19 at 14:27
• $f(h)$ is an operator that operates on diagonalisable matrices, so you really need to write $f(h)[\cdot]$ where, when you apply $f(h)[\cdot]$ instead of $\cdot$ you plug in a diagonalisable matrix. So I think $\mathbf{x}$ in the definition of $\|T\|_{op}$ is not a vector but a diagonalisable matrix. How is $M_2(\mathbb{R})$ defined? – them Aug 2 '19 at 14:30
• Unfortunately there weren't many details... I think it's just the space of all 2x2 real matrices. I'm still very confused as to how to derive a useful expression for $||f(h)-f(g)||$. The question did give specific matrices for $A$, $U$ and $D$ though - but I thought the function shouldn't depend on those matrices (unless the diagonal matrix was something silly like the zero matrix), so I left them out. – BaroqueFreak Aug 2 '19 at 14:34
• Well, another possibility is that you assume that $A$ is fixed. In this case your definitions make sense. And then you really need to write $f_A$, to show that for each $A$ you get a different map. And then show that for any $A$, $f_A$ is a continuous map. In which case explicitly $\|f_A(h)- f_A(g)\|_{op}=\sup_{\| \mathbf{x}\|=1}\|\left(U^{-1} h(D) U - U^{-1} g(D) U\right) \mathbf{x}\|$ – them Aug 2 '19 at 14:35

You do not tell us which norm you use on $$\mathbb R^2$$. But since all all norms on $$\mathbb R^2$$ are equivalent, we may take $$\lVert \mathbf{x} \rVert = \lVert \mathbf{x} \rVert_\infty = \max(\lvert x_1 \rvert, \lvert x_2 \rvert)$$.

Since the operator norm is submultiplicative (see e.g. show operator norm submultiplicative), we get

$$\lVert f(g) - f(h) \rVert_{op} = \lVert U^{-1}g(D) U - U^{-1}h(D) U \rVert_{op} = \lVert U^{-1}(g(D) - h(D)) U \rVert_{op} \\ \le \lVert U^{-1} \rVert_{op} \cdot \lVert g(D) - h(D) \rVert_{op} \cdot \lVert U \rVert_{op} = c \lVert g(D) - h(D) \rVert_{op} = c \lVert (g - h)(D) \rVert_{op} .$$ With $$u = g - h$$ we get $$\lVert u(D) \rVert_{op} = \sup_{\lVert \mathbf{x}\rVert=1}\lVert u(D)\mathbf{x} \rVert = \sup_{\lVert \mathbf{x}\rVert=1}\lVert \begin{pmatrix} u(d_{11})x_1\\ u(d_{22})x_2 \end{pmatrix} \rVert = \sup_{\lVert \mathbf{x}\rVert=1}\max( \lvert u(d_{11})x_1 \rvert, \lvert u(d_{22})x_2 \rvert) \\ = \max( \lvert u(d_{11}) \rvert, \lvert u(d_{22}) \rvert) \le \sup_{t \in [0,2]} \lvert u(t) \rvert = \lVert u \rVert_\infty$$ which finishes the proof.