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?

  • $\begingroup$ 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 $\|f(h) - f(g)\|_{op} < \varepsilon$ $\endgroup$ – them Aug 2 '19 at 14:26
  • $\begingroup$ What is $\lVert f(h)-f(g)\rVert$ explicitly? $\endgroup$ – BaroqueFreak Aug 2 '19 at 14:27
  • $\begingroup$ $f(h)$ 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? $\endgroup$ – them Aug 2 '19 at 14:30
  • $\begingroup$ 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. $\endgroup$ – BaroqueFreak Aug 2 '19 at 14:34
  • $\begingroup$ 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 $\|f_A(h)- f_A(g)\|_{op}=\sup_{\| \mathbf{x}\|=1}\|\left(U^{-1} h(D) U - U^{-1} g(D) U\right) \mathbf{x}\|$ $\endgroup$ – 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.

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