mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

1. a) Let $\mathrm{X}=\mathrm{C}[0,1]$$\mathrm{X}=\mathrm{C}[0,1]$X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]$\mathrm{X}=\mathrm{C}[0,1]$. Define $d:X\times X\to \mathbf{R}$$d:X\times X\to \mathbf{R}$d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}$d:X\times X\to \mathbf{R}$ by $\mathrm{d}(\mathrm{f},\mathrm{g})={\int }_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$$\mathrm{d}(\mathrm{f},\mathrm{g})={\int }_0^1 |\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}$\mathrm{d}(\mathrm{f},\mathrm{g})={\int }_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$ where the integral is the Riemann integral. Show that $d$$d$dd$d$ is a metric on $X$$X$XX$X$. Find $d(f,g)$$d(f,g)$d(f,g)d(f, g)$d(f,g)$ where $f(x)=4x$$f(x)=4x$f(x)=4xf(x)=4 x$f(x)=4x$ and $g(x)=x^3,x\in [0,1]$$g(x)=x^3,x\in [0,1]$g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]$g(x)=x^3,x\in [0,1]$.

1. Non-negativity: $d(f,g)\ge 0$$d(f,g)\ge 0$d(f,g) >= 0d(f, g) \geq 0$d(f,g)\ge 0$
2. Identity of indiscernibles: $d(f,g)=0$$d(f,g)=0$d(f,g)=0d(f, g) = 0$d(f,g)=0$ if and only if $f=g$$f=g$f=gf = g$f=g$
3. Symmetry: $d(f,g)=d(g,f)$$d(f,g)=d(g,f)$d(f,g)=d(g,f)d(f, g) = d(g, f)$d(f,g)=d(g,f)$
4. Triangle inequality: $d(f,h)\le d(f,g)+d(g,h)$$d(f,h)\le d(f,g)+d(g,h)$d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)$d(f,h)\le d(f,g)+d(g,h)$

1. Non-negativity:
   For any $f,g\in X$$f,g\in X$f,g in Xf, g \in X$f,g\in X$, the absolute value function $|\cdot |$$|\cdot |$|*||\cdot|$|\cdot |$ ensures that $|f(t)-g(t)|\ge 0$$|f(t)-g(t)|\ge 0$|f(t)-g(t)| >= 0|f(t) – g(t)| \geq 0$|f(t)-g(t)|\ge 0$ for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t \in [0, 1]$t\in [0,1]$. Therefore, the integral ${\int }_0^1|f(t)-g(t)|dt$${\int }_0^1 |f(t)-g(t)|dt$int_(0)^(1)|f(t)-g(t)|dt\int_0^1 |f(t) – g(t)| \, dt${\int }_0^1|f(t)-g(t)|dt$ is also non-negative. Hence, $d(f,g)\ge 0$$d(f,g)\ge 0$d(f,g) >= 0d(f, g) \geq 0$d(f,g)\ge 0$.
2. Identity of indiscernibles:
   If $f=g$$f=g$f=gf = g$f=g$, then $f(t)-g(t)=0$$f(t)-g(t)=0$f(t)-g(t)=0f(t) – g(t) = 0$f(t)-g(t)=0$ for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t \in [0, 1]$t\in [0,1]$, so $d(f,g)={\int }_0^1|f(t)-g(t)|dt={\int }_0^{1}0dt=0$$d(f,g)={\int }_0^1 |f(t)-g(t)|dt={\int }_0^1 0dt=0$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 0 \, dt = 0$d(f,g)={\int }_0^1|f(t)-g(t)|dt={\int }_0^{1}0dt=0$.
   Conversely, if $d(f,g)=0$$d(f,g)=0$d(f,g)=0d(f, g) = 0$d(f,g)=0$, then ${\int }_0^1|f(t)-g(t)|dt=0$${\int }_0^1 |f(t)-g(t)|dt=0$int_(0)^(1)|f(t)-g(t)|dt=0\int_0^1 |f(t) – g(t)| \, dt = 0${\int }_0^1|f(t)-g(t)|dt=0$. Since the integrand is non-negative, it must be zero almost everywhere, implying that $f(t)=g(t)$$f(t)=g(t)$f(t)=g(t)f(t) = g(t)$f(t)=g(t)$ for almost all $t\in [0,1]$$t\in [0,1]$t in[0,1]t \in [0, 1]$t\in [0,1]$. Since $f$$f$ff$f$ and $g$$g$gg$g$ are continuous, they must be equal everywhere on $[0,1]$$[0,1]$[0,1][0, 1]$[0,1]$, so $f=g$$f=g$f=gf = g$f=g$.
3. Symmetry:
   By the properties of the absolute value function, $|f(t)-g(t)|=|g(t)-f(t)|$$|f(t)-g(t)|=|g(t)-f(t)|$|f(t)-g(t)|=|g(t)-f(t)||f(t) – g(t)| = |g(t) – f(t)|$|f(t)-g(t)|=|g(t)-f(t)|$ for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t \in [0, 1]$t\in [0,1]$. Therefore, $d(f,g)={\int }_0^1|f(t)-g(t)|dt={\int }_0^1|g(t)-f(t)|dt=d(g,f)$$d(f,g)={\int }_0^1 |f(t)-g(t)|dt={\int }_0^1 |g(t)-f(t)|dt=d(g,f)$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)|g(t)-f(t)|dt=d(g,f)d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 |g(t) – f(t)| \, dt = d(g, f)$d(f,g)={\int }_0^1|f(t)-g(t)|dt={\int }_0^1|g(t)-f(t)|dt=d(g,f)$.
4. Triangle inequality:
   For any $f,g,h\in X$$f,g,h\in X$f,g,h in Xf, g, h \in X$f,g,h\in X$ and for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t \in [0, 1]$t\in [0,1]$, we have $|f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)|\le |f(t)-g(t)|+|g(t)-h(t)|$$|f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)|\le |f(t)-g(t)|+|g(t)-h(t)|$|f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)| <= |f(t)-g(t)|+|g(t)-h(t)||f(t) – h(t)| = |f(t) – g(t) + g(t) – h(t)| \leq |f(t) – g(t)| + |g(t) – h(t)|$|f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)|\le |f(t)-g(t)|+|g(t)-h(t)|$ by the triangle inequality for real numbers. Integrating both sides over $[0,1]$$[0,1]$[0,1][0, 1]$[0,1]$ gives:
   $${\int }_0^1|f(t)-h(t)|dt\le {\int }_0^1(|f(t)-g(t)|+|g(t)-h(t)|)dt={\int }_0^1|f(t)-g(t)|dt+{\int }_0^1|g(t)-h(t)|dt=d(f,g)+d(g,h)$$$${\int }_0^1 |f(t)-h(t)|dt\le {\int }_0^1 (|f(t)-g(t)|+|g(t)-h(t)|)dt={\int }_0^1 |f(t)-g(t)|dt+{\int }_0^1 |g(t)-h(t)|dt=d(f,g)+d(g,h)$$int_(0)^(1)|f(t)-h(t)|dt <= int_(0)^(1)(|f(t)-g(t)|+|g(t)-h(t)|)dt=int_(0)^(1)|f(t)-g(t)|dt+int_(0)^(1)|g(t)-h(t)|dt=d(f,g)+d(g,h)\int_0^1 |f(t) – h(t)| \, dt \leq \int_0^1 (|f(t) – g(t)| + |g(t) – h(t)|) \, dt = \int_0^1 |f(t) – g(t)| \, dt + \int_0^1 |g(t) – h(t)| \, dt = d(f, g) + d(g, h)$${\int }_0^1|f(t)-h(t)|dt\le {\int }_0^1(|f(t)-g(t)|+|g(t)-h(t)|)dt={\int }_0^1|f(t)-g(t)|dt+{\int }_0^1|g(t)-h(t)|dt=d(f,g)+d(g,h)$$
   Thus, $d(f,h)\le d(f,g)+d(g,h)$$d(f,h)\le d(f,g)+d(g,h)$d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)$d(f,h)\le d(f,g)+d(g,h)$.

1. For $0\le x\le 2$$0\le x\le 2$0 <= x <= 20 \leq x \leq 2$0\le x\le 2$, $4x\ge x^3$$4x\ge x^3$4x >= x^(3)4x \geq x^3$4x\ge x^3$, so $|4x-x^3|=4x-x^3$$|4x-x^3|=4x-x^3$|4x-x^(3)|=4x-x^(3)|4x – x^3| = 4x – x^3$|4x-x^3|=4x-x^3$.
2. For $2<x\le 1$$2<x\le 1$2 < x <= 12 < x \leq 1$2<x\le 1$, $4x<x^3$$4x<x^3$4x < x^(3)4x < x^3$4x<x^3$, so $|4x-x^3|=x^3-4x$$|4x-x^3|=x^3-4x$|4x-x^(3)|=x^(3)-4x|4x – x^3| = x^3 – 4x$|4x-x^3|=x^3-4x$.