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Gradient Descent $(12$ points$)$

$(2$ points$)$ Let $f$:$RR$ be a convex and $f({}^{**})=mi{n}_{}f().$ Prove that $f^{\text{'}}()0$ if $<msup\; id="EditorElement\_7404220"><mrow><mi\; id="EditorElement\_7404211"></mi></mrow><mrow\; id="EditorElement\_7404219"><mi\; id="EditorElement\_7404214">*</mi><mi\; id="EditorElement\_7404216">*</mi></mrow></msup>$ and $f^{\text{'}}()0$ if $>{}^{**}.$ Hint: You may want to use the alternative definition of convexity that $f(y)f(x)+f^{\text{'}}(x)(y-x).$

$(2$ points$)$ We say a vec$()$ is a local minimum for a function $f$:$R^{d}R$ if there exists $lon>0$ such that ${}^{\text{'}}||(vec())-$vecvec$\frac{({)}^{(i)}}{b}ar()f(\frac{?}{\text{b}\text{a}\text{r}}())f(vec({)}^{**})+lonff$:$R^{d}RG||gradf()|{|}_{2}G{}^{(i+1)}={}^{(i)}-gradf({}^{(i)})||{}^{(i+1)}-{}^{(i)}|{|}_{2}Gt=\frac{R^2{G}^2}{lo{n}^2}=\frac{R}{G\sqrt[\phantom{2}]{t}}ii+1f({}^{(i+1)})-f({}^{(i)})S{}^{(i+1)}=P_S({}^{\text{o}\text{u}\text{t}}){}^{\text{o}\text{u}\text{t}}={}^{(i)}-gradf({}^{(i)})f(vec())$ for all ${}^{\text{'}}$ such that $||(vec())-$vec Show that $iffis$ convex then there can $at$ most one local minimum vec$({)}^{**}.$

$(2$ points$)In$ our gradient descent analysis $we$ used the fact $\frac{1}{t}{\displaystyle \underset{i=1}{\overset{t}{}}}f(vec({)}^{(i)})f(vec({}^{**}))+lon$ implies that $f(hat())f(vec({)}^{**})+lon$ for the best iterate hat$()=argmi{n}_{\text{v}\text{e}\text{c}({)}^{(1)},\text{d}\text{o}\text{t}\text{s}\text{,}\text{v}\text{e}\text{c}({)}^{(t)}}f(vec({}^{(i)})).$ Prove that $ifwe$ instead set $\frac{?}{\text{b}\text{a}\text{r}}()=\frac{1}{t}{\displaystyle \underset{i=1}{\overset{t}{}}}$vec$({)}^{(i)}(i.e.,\frac{?}{\text{b}\text{a}\text{r}}()is$ the average iterate$)$ then $we$ also have $f(\frac{?}{\text{b}\text{a}\text{r}}())f(vec({)}^{**})+lon.$ Hint: Use that $fis$ convex.

Let $f$:$R^{d}RbeaG-$Lipschitz function, $i.e.,||gradf()|{|}_{2}G$ for all $.$

$(a)(2$ points$)If{}^{(i+1)}={}^{(i)}-gradf({}^{(i)}),$ give $an$ upper bound $on||{}^{(i+1)}-{}^{(i)}|{|}_{2}in$ terms $of$ and $G.$

$(b)(2$ points$)In$ our fixed step size gradient algorithm $we$ set $t=\frac{R^2{G}^2}{lo{n}^2}$ and $=\frac{R}{G\sqrt[\phantom{2}]{t}}.$ Under these settings, what $is$ the worst case increase $in$ function value from step $ito$ step $i+1?$ I.$e.,$ give $an$ upper bound $onf({}^{(i+1)})-f({}^{(i)}).$ Does this make intuitive sense? Hint: Use part $(a).$

$(c)(2$ points$)$ Consider the case $of$ projected gradient descent over a convex set $S.So{}^{(i+1)}=P_S({}^{\text{o}\text{u}\text{t}})$ for ${}^{\text{o}\text{u}\text{t}}={}^{(i)}-gradf({}^{(i)}).$ Show that the bounds $of(a),(b)$ still hold.